All Solutions
Page 66: Assessment
begin{center}
begin{tabular}{|l|c|}
hline
House to Friend’s House & $2.1~mathrm{km}$ \ hline
House to Grocery Store & $4.3~mathrm{km}$ \ hline
end{tabular}
end{center}
$$
d_{1} = 4.3~mathrm{km}
$$
$$
d_{2} = 4.3~mathrm{km} + 2.1~mathrm{km} = 6.4~mathrm{km}
$$
$$
d_{3} = 2.1~mathrm{km}
$$
$$
begin{align*}
d &= d_{1} + d_{2} + d_{3} \
&= left( 4.3~mathrm{km} right) + left( 6.4~mathrm{km} right) + left( 2.1~mathrm{km} right)\
d &= boxed{12.8~mathrm{km}}
end{align*}
$$
12.8~mathrm{km}
$$
$$
begin{align*}
Delta x &= x_text{f} – x_text{i} \
&= 0 – left( 8.0~mathrm{m} right) \
Delta x &= boxed{ -8.0~mathrm{m} }
end{align*}
$$
$$
begin{align*}
d &= d_{1} + d_{2} \
&= left| 14~mathrm{m} – 8.0~mathrm{m} right| + left| 0 – 14~mathrm{m} right| \
&= 6~mathrm{m} + 14~mathrm{m} \
d &= boxed{20~mathrm{m}}
end{align*}
$$
d = 20~mathrm{m}
$$
The distance is always non-negative. Also, since the object moved through a path, the distance cannot be 0, and it is therefore $textbf{positive}$.
In a round-trip, the initial and final positions are the same, hence the displacement must be $textbf{zero}$.
The displacement is $textbf{the same}$, since they have the same final and initial positions.
The distance is $textbf{not the same}$, since the dog’s short trips would add to its distance.
In this part, we find the distance. We simply add the distance traveled for the two parts of the trip.
$$
begin{align*}
d &= d_{1} + d_{2} \
&= left| x_{2} – x_{0} right| + left| x_{1} – x_{2} right| \
&= 6.2~mathrm{m} + 1.2~mathrm{m} \
d&= boxed{7.4~mathrm{m}}
end{align*}
$$
In this part, we find the displacement. We only check the final position and initial position.
$$
begin{align*}
Delta x &= x_{1} – x_{0} \
&= 5.0~mathrm{m} – 0~mathrm{m} \
Delta x &= boxed{ 5.0~mathrm{m} }
end{align*}
$$
item [a)] $d = 7.4~mathrm{m}$
item [b)] $Delta x = 5.0~mathrm{m}$
end{enumerate}
The distance total distance is simply the sum of the distances for each part of the path.
$$
begin{align*}
d &= d_{1} + d_{2} \
&= left| x_{1} – x_{0} right| + left| x_{2} – x_{1} right| \
&= 22~mathrm{m} + 29.5~mathrm{m} \
&= 51.5~mathrm{m} \
d&= boxed{52~mathrm{m}}
end{align*}
$$
The displacement depends only on the final and initial position.
$$
begin{align*}
Delta x &= x_{2} – x_{0} \
&= -7.5~mathrm{m} – 0~mathrm{m} \
Delta x &= boxed{ -7.5~mathrm{m} }
end{align*}
$$
item [a)] $d = 52~mathrm{m}$
item [b)] $Delta x = -7.5~mathrm{m}$
end{enumerate}
We see that the train moves in opposite direction, so that means that the displacement will be lower since it moves closer to the initial position, but the distance only gets larger. Hence, the distance is $textbf{greater than}$ the displacement.
In this part, we calculate the distance covered by the train. We need to add the distances
$$
begin{align*}
d &= d_{1} + d_{2} \
&= left| Delta x_{1} right| + left| Delta x_{2} right| \
&= 5.9~mathrm{km} + 3.8~mathrm{km} \
d &= boxed{ 9.7~mathrm{km} }
end{align*}
$$
The net displacement is the sum of the individual displacements, since the individual displacements already account for the directions.
$$
begin{align*}
Delta x &= Delta x_{1} + Delta x_{2} \
&= 5.9~mathrm{km} – 3.8~mathrm{km} \
Delta x &= boxed{ 2.1~mathrm{km} }
end{align*}
$$
$$
begin{align*}
y &= vt \
&= left( 4.6~mathrm{m/{color{#c34632} s}} right) left( 1.4~mathrm{color{#c34632} s} right) \
&= 6.44~mathrm{m} \
y &= boxed{ 6.4~mathrm{m} }
end{align*}
$$
6.4~mathrm{m}
$$
$$
begin{align*}
d &= vt \
&= left( 340~mathrm{m/{color{#c34632} s}} right) left( 3.50~mathrm{color{#c34632} s} right) \
&= 1190~mathrm{m} = 1.19~mathrm{km} \
d &= boxed{ 1200~mathrm{m} = 1.2~mathrm{km} }
end{align*}
$$
d = 1200~mathrm{m} = 1.2~mathrm{km}
$$
In this part, we find the distance traveled if the red kangaroo hops for $t = 3.2~mathrm{min}$. We have
$$
begin{align*}
d &= vt \
&= left( 65~mathrm{km/h} right) left( 3.2~mathrm{min} right) \
&= left( 65~mathrm{km/{color{#c34632}hr }} right) left( frac{1~mathrm{color{#c34632}hr}}{60~mathrm{color{#4257b2} min}} right) left( 3.2~mathrm{color{#4257b2}min} right) \
&= 3.466667~mathrm{km} \
d &= boxed{ 3.5~mathrm{km} }
end{align*}
$$
For this part, we find it takes for the kangaroo to hop $d = 0.25~mathrm{km}$.
$$
begin{align*}
t &= frac{d}{v} \
&= frac{0.25~mathrm{color{#c34632}km}}{65~mathrm{{color{#c34632}km}/h}} \
&= 3.84615 times 10^{-3}~mathrm{hr} \
t &= boxed{3.8 times 10^{-3}~mathrm{hr}}
end{align*}
$$
item [a)] $d = 3.5~mathrm{km}$
item [b)] $t = 3.8 times 10^{-3}~mathrm{hr}$
end{enumerate}
$$
begin{align*}
v_text{ave} &= frac{d_text{total}}{t_text{total}} \
&= frac{d_{1} + d_{2}}{t_{1} + t_{2}} \
&= frac{v_{1}t_{1} + v_{2}t_{2}}{t_{1} + t_{2}} \
&= frac{ left( 0.060~mathrm{m/s} right) left( 1.2~mathrm{min} right) + left( 13~mathrm{m/s} right) left( 1.2~mathrm{min} right) }{1.2~mathrm{min} + 1.2~mathrm{min}} \
&= frac{ 0.060~mathrm{m/s} + 13~mathrm{m/s} }{2} \
&= 6.53~mathrm{m/s} \
v_text{ave} &= boxed{ 6.5~mathrm{m/s} }
end{align*}
$$
v_text{ave} = 6.5~mathrm{m/s}
$$
$$
begin{align*}
v_text{av} &= frac{x_text{f} – x_text{i}}{Delta t} \
implies Delta t &= frac{x_text{f} – x_text{i}}{v_text{av}} \
&= frac{0 – 50.0~mathrm{m}}{-1.50~mathrm{m/s}} \
Delta t &= boxed{ 33.3~mathrm{s} }
end{align*}
$$
t = 33.3~mathrm{s}
$$
First we convert the time into seconds
$$
begin{align*}
t &= 80~mathrm{color{#c34632}days} left( frac{24~mathrm{color{#4257b2}hours}}{1~mathrm{color{#c34632}day}} right) left( frac{60~mathrm{color{#19804f}min}}{1~mathrm{color{#4257b2}hour}} right) left( frac{60~mathrm{s}}{1~mathrm{color{#19804f}min}} right) \
t &= 6912000~mathrm{s}
end{align*}
$$
$$
begin{align*}
s &= frac{d}{t} \
&= frac{40075~mathrm{color{#c34632}km}}{6912000~mathrm{s}} left( frac{1000~mathrm{m}}{1~mathrm{color{#c34632}km}} right) \
&= 5.797887~mathrm{m/s} \
s &= boxed{ 5.8~mathrm{m/s} }
end{align*}
$$
Since the path is a roundtrip, the traveler stops at the initial position. Therefore, the displacement is $0$, and the velocity $v$ must be
$$
boxed{ v = 0 }
$$
item [a)] $5.8~mathrm{m/s}$
item [b)] $0~mathrm{m/s}$
end{enumerate}
Train A travels north with speed 10 m/s. Its velocity must be
$$
v_{A} = +10~mathrm{m/s}
$$
Train B moves south and covers 900 m in 1 min = 60 s. Its speed is $dfrac{900~mathrm{m}}{60~mathrm{s}} = 15~mathrm{m/s}$ and its velocity is
$$
v_{B} = -15~mathrm{m/s}
$$
Train C moves south and has twice the speed of train A. Train C must have a speed of 20 m/s, and has velocity
$$
v_{C} = -20~mathrm{m/s}
$$
Train D travels north and covers 25 m in 2 s. Its speed is $dfrac{24~mathrm{m}}{2~mathrm{s}} = 12~mathrm{m/s}$ and its velocity is
$$
v_{D} = +12~mathrm{m/s}
$$
$$
boxed{ v_{C} < v_{B} < v_{A} < v_{D} }
$$
v_{C} < v_{B} < v_{A} < v_{D}
$$
mathrm{m/s}
$$
$$
begin{align*}
s &= frac{d}{t} \
&= frac{100~mathrm{m}}{9.58~mathrm{s}} \
&= 10.43841~mathrm{m/s} \
s &= boxed{ 10.4~mathrm{m/s} }
end{align*}
$$
$$
begin{align*}
s &= frac{d}{t} \
&= frac{100~mathrm{color{#c34632}m}}{9.58~mathrm{color{#4257b2}s}} left( frac{1~mathrm{km}}{1000~mathrm{color{#c34632}m}} right) left( frac{3600~mathrm{color{#4257b2}s}}{1~mathrm{h}} right)\
&= 37.57829mathrm{km/h} \
s &= boxed{ 37.6~mathrm{km/h} }
end{align*}
$$
10.4~mathrm{m/s} = 37.6~mathrm{km/h}
$$
$$
begin{align*}
t &= frac{d_text{total}}{s} \
&= frac{2d}{s} \
&= frac{2 left( 3.8 times 10^{8}~mathrm{color{#c34632}m} right)}{300000000~mathrm{{color{#c34632}m}/s}} \
&= 2.53333~mathrm{s} \
t &= boxed{2.5~mathrm{s}}
end{align*}
$$
2.5~mathrm{s}
$$
The distance traveled is te same for both speeds. However, the time for the first speed is greater than the time for the second speed. The average speed must be closer to the slower speed, which is $20~mathrm{m/s}$. The average speed is less than $25~mathrm{m/s}$, since at that speed, the times for the two rates are equal, but they are not.
To find the average speed, we have
$$
begin{align*}
v_text{av} &= frac{d_text{total}}{t_text{total}} \
&= frac{d_{1} + d_{2}}{frac{d_{1}}{v_{1}} + frac{d_{2}}{v_{2}}} \
&= frac{ 2~mathrm{km} + 2~mathrm{km} }{frac{2~mathrm{km}}{20~mathrm{m/s}} + frac{2~mathrm{km}}{30~mathrm{m/s}}} \
v &= boxed{ 24~mathrm{m/s} }
end{align*}
$$
$$
begin{align*}
v_text{av} &= frac{d_text{total}}{t_text{total}} \
&= frac{d_{1} + d_{2}}{frac{d_{1}}{v_{1}} + frac{d_{2}}{v_{2}}} \
&= frac{ 1~mathrm{km} + 1~mathrm{km} }{frac{1~mathrm{km}}{10~mathrm{m/s}} + frac{1~mathrm{km}}{30~mathrm{m/s}}} \
v &= boxed{ 15~mathrm{m/s} }
end{align*}
$$
The speed is indeed less than $20~mathrm{m/s}$.
The speeds are the corresponding slopes for the time periods. The graph would be
From the graph, we see that
$$
begin{align*}
xleft( t = 2~mathrm{s} right) &= 4.0~mathrm{m} \
xleft( t = 11~mathrm{s} right) &= 11~mathrm{m}
end{align*}
$$
We first get the speed, which is the magnitude of the slope for the different segments. From the graph, we see that the steepest slope is segment $B$, followed by $A$, then $C$. The order of increasing speed must be
$$
boxed{ C < A < B }
$$
In this part, we rank the velocities from the most negative to the most positive. Both segments $A$ and $C$ have negative slopes, with the slope of $A$ begin steeper. Segment $B$ has a positive slope. The rank of increasing velocity must be
$$
boxed{ A < C < B }
$$
item [a)] $C < A < B$
item [b)] $A < C < B$
end{enumerate}
For this part, we compare the magnitude of the slopes. From the graph, the order from least steep to steepest is
$$
boxed{ C < B < A < D }
$$
For this part, we also take into account if the slope is negative and positive. The rank of increasing velocity is
$$
boxed{ D < C < B < A }
$$
item [a)] $C < B < A < D$
item [b)] $D < C < B < A$
end{enumerate}
$$
begin{align*}
v_text{av} &= frac{Delta x}{Delta t} \
&= frac{2~mathrm{m} – 3~mathrm{m}}{5~mathrm{s} – 0} \
v_text{av}&= boxed{ -0.2~mathrm{m/s} }
end{align*}
$$
v_text{av} = -0.2~mathrm{m/s}
$$
For this part, we check which segment has the least steep slope. From the graph, this is segment
$$
boxed{B}
$$
To find this speed, we get the magnitude of the slope.
$$
begin{align*}
s &= left| frac{Delta x}{Delta t} right| \
&= left| frac{-2~mathrm{m} – left( 1~mathrm{m} right)}{4~mathrm{s} – 2~mathrm{s}} right| \
s &= boxed{ 1.5~mathrm{m/s} }
end{align*}
$$
item [a)] $B$
item [b)] $1.5~mathrm{m/s}$
end{enumerate}
The average velocity is the ratio of the displacement and time.
$$
begin{align*}
v_text{av} &= frac{Delta x}{Delta t} \
&= frac{5~mathrm{m} – left( -3~mathrm{m} right)}{7~mathrm{s} – 0} \
&= 1.142857~mathrm{m/s} \
v_text{av}&= boxed{ 1.1~mathrm{m/s} }
end{align*}
$$
v_text{av} = 1.1~mathrm{m/s}
$$
$$
x_text{f} = 8.3~mathrm{m} + left( 2.2~mathrm{m/s} right)t
$$
We find its initial position and speed.
Comparing the given equation to the formula
$$
begin{align*}
x_text{f} &= x_text{i} + vt tag{1}
end{align*}
$$
we see that
$$
boxed{ x_text{i} = 8.3~mathrm{m} }
$$
Still using equation (1), we see that
$$
boxed{ v = 2.2~mathrm{m/s} }
$$
item [a)] $x_text{i} = 8.3~mathrm{m}$
item [b)] $v = 2.2~mathrm{m/s}$
end{enumerate}
In this part, we first find the velocity of the bowling ball. We have
$$
begin{align*}
v &= frac{v_text{f} – v_text{i}}{t} \
&= frac{ 7.8~mathrm{m} – 1.6~mathrm{m} }{3.1~mathrm{s}} \
v &= 2.0~mathrm{m/s}
end{align*}
$$
$$
begin{align*}
x_text{f} &= x_text{i} + vt \
x_text{f} &=boxed{ 1.6~mathrm{m} + left( 2.0~mathrm{m/s} right)t } tag{1}
end{align*}
$$
For this part, $x_text{f} = 8.6~mathrm{m}$. We find the time $t$ using equation (1).
$$
begin{align*}
x_text{f} &= 1.6~mathrm{m} + left( 2.0~mathrm{m/s} right)t \
8.6~mathrm{m} &= 1.6~mathrm{m} + left( 2.0~mathrm{m/s} right)t \
implies t &= frac{8.6~mathrm{m} – 1.6~mathrm{m}}{2.0~mathrm{m/s}} \
t &= boxed{ 3.5~mathrm{s} }
end{align*}
$$
item [a)] $x_text{f} = 1.6~mathrm{m} + left( 2.0~mathrm{m/s} right)t$
item [b)] $t = 3.5~mathrm{s}$
end{enumerate}
For this part, we find compare the velocities of $A$ and $B$. We see that the position of $B$ becomes more positive faster than $A$, so we have
$$
boxed{ v_{A} < v_{B} }
$$
For this part, we compare the initial position of the two objects. From the graph, $x_text{i, A} = 12.5~mathrm{m}$ and $x_text{i, B} = 5.00~mathrm{m}$, hence
$$
boxed{ x_text{i, A} > x_text{i, B} }
$$
For this part, we compare $x_{A}$ and $x_{B}$ for time $t = 0$ to $t = 1~mathrm{s}$. From the graph, we easily see that
$$
boxed{ x_{A} > x_{B} }
$$
item [a)] $v_{A} x_text{i, B}$
item [c)] $x_{A} > x_{B}$
end{enumerate}
$$
begin{align*}
x_text{f} &= x_text{i, fr} + v_text{fr}t_{1} \
&= 7.8~mathrm{m} + left( 2.3~mathrm{m/s} right) left( 2.2~mathrm{s} right) \
x_text{f} &= 12.86~mathrm{m}
end{align*}
$$
$$
begin{align*}
x_text{f} &= x_text{i, O} + v_text{O}t_{2} \
implies t_{2} &= frac{ x_text{f} – x_text{i, O} }{v_text{O}} \
&= frac{12.86~mathrm{m} – 0~mathrm{m}}{4.2~mathrm{m/s}} \
&= 3.06190~mathrm{s} \
t_{2} &= 3.1~mathrm{s}
end{align*}
$$
The observer catches up with the friend at time $t_{2} = 3.1~mathrm{s}$.
t_{2} = 3.1~mathrm{s}
$$
$$
begin{align*}
x_text{f} &= x_text{i} + v_text{f}t
end{align*}
$$
$$
begin{align*}
x_text{f, O} &= boxed{left( 26~mathrm{m/s} right)t}
end{align*}
$$
For the truck, it is
$$
begin{align*}
x_text{f, tr} &= boxed{ 420~mathrm{m} + left( -31~mathrm{m/s} right)t}
end{align*}
$$
Below is the graph of their position time graphs.
To find the time in which the pass one another, we equate their position-time equation.
$$
begin{align*}
x_text{f, 0} &= x_text{f, tr} \
left( 26~mathrm{m/s} right)t &= 420~mathrm{m} + left( -31~mathrm{m/s} right)t \
left( 26~mathrm{m/s} – left[ -31~mathrm{m/s} right] right)t &= 420~mathrm{m} \
implies t &= frac{420~mathrm{m}}{26~mathrm{m/s} + 31~mathrm{m/s}} \
&= 7.3684~mathrm{s} \
t &= boxed{ 7.4~mathrm{s} }
end{align*}
$$
$$
begin{align*}
x_{1} &= -4.0~mathrm{m} + left( 2.5~mathrm{m/s} right)t \
x_{2} &= 6.7~mathrm{m} + left( -3.5~mathrm{m/s} right)t
end{align*}
$$
We compare their speeds.
$$
begin{align*}
x_{1} &= -4.0~mathrm{m} + left( 2.5~mathrm{m/s} right)t \
x_{2} &= 6.7~mathrm{m} + left( -3.5~mathrm{m/s} right)t
end{align*}
$$
$$
begin{align*}
d &= left| x_{1} – x_{2} right| \
&= left| left[ -4.0~mathrm{m} + left( 2.5~mathrm{m/s} right)t right] – left[ 6.7~mathrm{m} + left( -3.5~mathrm{m/s} right)t right] right| \
d &= left| -10.7~mathrm{m} + left( 6.0~mathrm{m/s} right)t right| tag{1}
end{align*}
$$
$$
begin{align*}
d &= left| -10.7~mathrm{m} + left( 6.0~mathrm{m/s} right)t right| \
&= left| -10.7~mathrm{m} + left( 6.0~mathrm{m/s} right)left( 0 right) right| \
d &= boxed{ 10.7~mathrm{m} }
end{align*}
$$
$$
begin{align*}
d &= left| -10.7~mathrm{m} + left( 6.0~mathrm{m/s} right)t right| \
&= left| -10.7~mathrm{m} + left( 6.0~mathrm{m/s} right)left( 1~mathrm{s} right) right| \
d &= boxed{ 4.7~mathrm{m} }
end{align*}
$$
$$
x = 3.0~mathrm{m} + left( -5.0~mathrm{m/s} right)t
$$
We find its position at time $t = 1.5~mathrm{s}$
$$
begin{align*}
x &= 3.0~mathrm{m} + left( -5.0~mathrm{m/s} right)t \
&= 3.0~mathrm{m} + left( -5.0~mathrm{m/s} right) left( 1.5~mathrm{s} right) \
x &= boxed{ -4.5~mathrm{m} }
end{align*}
$$
-4.5~mathrm{m}
$$
$$
begin{align*}
x_{1} &= -4.0~mathrm{m} + left( 1.5~mathrm{m/s} right)t \
x_{2} &= 8.8~mathrm{m} + left( -2.5~mathrm{m/s} right)t
end{align*}
$$
We compare their speeds and find the time in which they collide.
The speed is the magnitude of the velocity. For bumper 1, the speed is $s_{1} = 1.5~mathrm{m/s}$ and for bumper 2, the speed is $s_{2} = 2.5~mathrm{m/s}$. We see that
$$
boxed{ s_{2} > s_{1} }
$$
To find the time in which they collide, we equate their positions.
$$
begin{align*}
x_{1} &= x_{2} \
-4.0~mathrm{m} + left( 1.5~mathrm{m/s} right)t &= 8.8~mathrm{m} + left( -2.5~mathrm{m/s} right)t \
left( 1.5~mathrm{m/s} – left[ -2.5~mathrm{m/s} right] right)t &= 8.8~mathrm{m} – left[ -4.0~mathrm{m} right] \
implies t &= frac{8.8~mathrm{m} + 4.0~mathrm{m}}{1.5~mathrm{m/s} + 2.5~mathrm{m/s}} \
t &= boxed{ 3.2~mathrm{s} }
end{align*}
$$
item [a)] $s_{2} > s_{1}$
item [b)] $t = 3.2~mathrm{s}$
end{enumerate}
On the other hand, $textbf{displacement is the change in position.}$ It is also a vector, and the sign of the displacement vector indicates the direction of motion. This mean that displacement can be positive, negative, and also zero.
Although these two quantities seem almost the same, they are very different.
Another example is when you take a walk around the block: $textbf{the distance travelled is certainly positive, but the displacement is zero because the initial and final position are the same;}$ there is no net change in position.
Two examples are given: the first being a situation when the magnitude of displacement is greater than zero, and the second one is a situation in which displacement is exactly zero.
Overall, there is no scenario in which the magnitude of your displacement is greater than the distance you cover.
On the other hand, $textbf{when you travel from home to school, there is a net change in position.}$ The distance you covered is nowhere near the distance the astronaut covers, but you have the greater displacement.
It goes without saying that you covered greater distance than your friend.
$$
begin{align*}
boxed{Delta x=x_{f}-x_{i}}\
end{align*}
$$
Even without plugging in the values, it’s easy to see that $textbf{you have a positive displacement, and your friend has a negative displacement.}$ The change in your position is in the positive direction (from $x_{i}=20 text{m}$ to $x_{f}=25 text{m}$), therefore, your displacement is positive. On the other hand, the change in your friend’s position was in the negative direction (from $x_{i}=35 text{m}$ to $x_{f}=30 text{m}$), which means that his displacement was negative.
$$
begin{align*}
d&=0.75 text{km}+0.60 text{km}+0.60 text{km}\
&=quadboxed{1.95 text{km}}\
end{align*}
$$
$$
begin{align*}
Delta x&=x_{f}-x_{i}\
&=0.75 text{km}-0 text{km}\
&=quadboxed{0.75 text{km}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{d=1.95 text{km}}\
textbf{(b)} quad &boxed{Delta x=0.75 text{km}}\
end{align*}
$$
$$
begin{align*}
d&=0.60 text{km}+0.35 text{km}+0.35 text{km}+0.60 text{km}+0.75 text{km}\
&=quadboxed{2.65 text{km}}\
end{align*}
$$
$$
begin{align*}
Delta x&=x_{f}-x_{i}\
&=0.75 text{km}-0 text{km}\
&=quadboxed{0.75 text{km}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{d=2.65 text{km}}\
textbf{(b)} quad &boxed{Delta x=0.75 text{km}}\
end{align*}
$$
The difference between distance and displacement is that former is the total length of travel, and the later is the net change in position.
$$
textbf{Solution:}
$$
Now sum the distances to find the total length travelled
$$
(0.60+0.35km)+(0.35+0.60+0.75km)=color{#4257b2} boxed{bf 2.65km}
$$
The difference between distance and displacement is that former is the total length of travel, and the later is the net change in position.
$$
textbf{Solution:}
$$
Now subtract $x_i$ from $x_f$ to compute the displacement
$$
Delta x = x_f-x_i=0.75-0.00km=color{#4257b2} boxed{bf 0.75km}
$$
Delta x =0.75km
$$
$$
begin{align*}
Delta x&=x_{f}-x_{i}\
6.2 text{m}&=x_{f}-4.5 text{m}\
\
rightarrow x_{f}&=6.2 text{m}+4.5 text{m}\
&=quadboxed{10.7 text{m}}\
end{align*}
$$
begin{align*}
boxed{x_{f}=10.7 text{m}}\
end{align*}
$$
textbf{Concept:}
$$
Displacement is defined as the difference between and object’s initial and final positions. With this fact determine your final position given your initial position and displacement.
$$
textbf{Solution:}
$$
Solve for $x_f$ from the displacement equation
$$
Delta x = x_f-x_i
$$
$$
Delta x + x_i=x_f
$$
$$
(6.2m)+(4.5m)=x_f=color{#4257b2} boxed{bf 10.7m}
$$
x_f=10.7m
$$
$$
begin{align*}
Delta x&=x_{f}-x_{i}\
-8.3 text{m}&=x_{f}-7.5 text{m}\
\
rightarrow x_{f}&=-8.3 text{m}+7.5 text{m}\
&=quadboxed{-0.8 text{m}}\
end{align*}
$$
begin{align*}
boxed{x_{f}=-0.8 text{m}}\
end{align*}
$$
textbf{Concept:}
$$
Displacement is defined as the difference between and object’s initial and final positions. With this fact determine your final position given your initial position and displacement.
$$
textbf{Solution:}
$$
Solve for $x_f$ from the displacement equation
$$
Delta x = x_f-x_i
$$
$$
Delta x + x_i=x_f
$$
$$
(-8.3m)+(7.5m)=x_f=color{#4257b2} boxed{bf -0.8m}
$$
x_f=-0.8m
$$
$$
begin{align*}
Delta x&=x_{f}-x_{i}\
-26 text{m}&=4.3 text{m}-x_{i}\
\
rightarrow x_{i}&=4.3 text{m}+26 text{m}\
&=30.3 text{m}\
&=quadboxed{30 text{m}}\
end{align*}
$$
When writing the final result, we used $textbf{the rule for addition and subtraction for significant figures.}$
begin{align*}
boxed{x_{i}=30 text{m}}\
end{align*}
$$
textbf{Concept:}
$$
Displacement is defined as the difference between and object’s initial and final positions. With this fact determine your final position given your initial position and displacement.
$$
textbf{Solution:}
$$
Solve for $x_f$ from the displacement equation
$$
Delta x = x_f-x_i
$$
$$
x_i=x_f-Delta x
$$
$$
x_i=(4.3m)-(-26m)=x_f=color{#4257b2} boxed{bf 30m}
$$
x_f=30m
$$
$$
begin{align*}
Delta x&=x_{f}-x_{i}\
17 text{m}&=-2.2 text{m}-x_{i}\
\
rightarrow x_{i}&=-2.2 text{m}-17 text{m}\
&=-19.2 text{m}\
&=quadboxed{-19 text{m}}\
end{align*}
$$
When writing the final result, we used $textbf{the rule for addition and subtraction for significant figures.}$
begin{align*}
boxed{x_{i}=-19 text{m}}\
end{align*}
$$
textbf{Concept:}
$$
For the train the definition of displacement remains the same and is defined as the difference between and objects’s initial and final positions. With this fact determine your final position given your initial position and displacement.
$$
textbf{Solution:}
$$
Solve for $x_i$ from the displacement equation by computing $x_i$ from the equation
$$
Delta x = x_f-x_i
$$
$$
x_i=x_f-Delta x
$$
$$
x_i=(-2.2m)-(17m)=color{#4257b2} boxed{bf -19.2m}
$$
x_i=-19.2m
$$
Distance is defined as the total length travelled, whereas displacement is the net change in position, and it takes into consideration the direction of motion.
Now that we’ve set the coordinate system, and repeated the definitions for distance and displacement, we can easily solve the problem.
$$
begin{align*}
&circquadboxed{d_{A}=5 text{m}}\
end{align*}
$$
$$
begin{align*}
circquadDelta x_{A}&=x_{f}-x_{i}\
&=0 text{m}-(-5 text{m})\
&=quadboxed{5 text{m}}\
end{align*}
$$
$$
begin{align*}
&circquadboxed{d_{B}=2 text{m}}\
end{align*}
$$
$$
begin{align*}
circquadDelta x_{B}&=x_{f}-x_{i}\
&=0 text{m}-2 text{m}\
&=quadboxed{-2 text{m}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{d_{A}=5 text{m}}\
&boxed{Delta x_{A}=5 text{m}}\
\
textbf{(b)} quad &boxed{d_{B}=2 text{m}}\
&boxed{Delta x_{B}=-2 text{m}}\
end{align*}
$$
The distance travelled by the ball is
$$
begin{align*}
d&=10 text{m}+2.5 text{m}+2.5 text{m}\
&=quadboxed{15 text{m}}\
end{align*}
$$
$$
begin{align*}
Delta x&=x_{f}-x_{i}\
&=10 text{m}-0 text{m}\
&=quadboxed{10 text{m}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{d=15 text{m}}\
textbf{(b)} quad &boxed{Delta x=10 text{m}}\
end{align*}
$$
The distance is related to the total length of travel whereas the displacement is net change of position.
$$
textbf{Solution:}
$$
Sum lengths to find the distance travelled:
$$
(10+2.5m)+2.5m=color{#4257b2} boxed{bf 15m}
$$
Get the difference of final and initial position to find the displacement
$$
textbf{Solution:}
$$
$$
Delta x = x_f-x_i=10-0m=color{#4257b2} boxed{bf 10m}
$$
Delta x=10m
$$
$$begin{align*}
d_{A}=15text{ m} + 100text{ m} + 15text{ m} = boxed{130 text{m}}\
end{align*}$$
The displacement in the horizontal direction is:
$$begin{align*}
Delta x_{A}&=x_{f}-x_{i}\
&=100 text{m}-0 text{m}\
&=boxed{100 text{m}}\
end{align*}$$
whereas the vertical displacement is zero because the jogger runs the same distances in opposite directions.
$$begin{align*}
d_{B}&=100 text{m}+30 text{m}+100 text{m}+30 text{m}\
&=boxed{260 text{m}}\
end{align*}$$
However, the displacement is zero: the jogger finishes the run at the same position they started; it’s around-trip.
$$begin{align*}
Delta x_{B}&=x_{f}-x_{i}\
&=boxed{0 text{m}}\
end{align*}$$
Another way to see this is to analyze each segment separately: first, the jogger moves $100 text{m}$ in the positive horizontal direction, then $30 text{m}$ in the positive vertical direction, after that again $100 text{m}$ but in the negative horizontal direction, and finally $30 text{m}$ in the negative vertical direction. The displacement is zero in both directions.
textbf{(a)} quad &d_{A}=130 text{m}\
&Delta x_{A}=100 text{m}\
\
textbf{(b)} quad &d_{B}=260 text{m}\
&Delta x_{B}=0 text{m}\
end{align*}$$
We already knew that the distance is the total length of travel, and the displacement is the net change in position. To calculate the distance traveled by the runner sum the distances along x and y axis, the runner first goes 15 m in the –y direction, then 100 m in the +x direction, and then 15 m in the +y direction.
**Calculation:**
Sum distances to find the total distance travelled.
$$15text{ m}+100text{ m}+15text{ m}=color{#4257b2} boxed{bf 130 m}$$
Now calculate the displacement:
$$Delta x = x_f-x_i=100text{ m}-0=color{#4257b2} boxed{bf 100 m}$$
We already knew that the distance is the total length of travel, and the displacement is the net change in position. To calculate the distance traveled by the runner sum the distances along x and y axis, the runner first goes 15 m in the –y direction, then 100 m in the +x direction, and then 15 m in the +y direction.
**Calculation:**
Sum distances to find the total distance travelled.
$$15text{ m}+100text{ m}+30text{ m}+100text{ m}+15text{ m}=color{#4257b2} boxed{bf 260 m}$$
Now calculate the displacement:
$$Delta x = x_f-x_i=0-0=color{#4257b2} boxed{bf 0 m}$$
$$d = 260text{ m and }Delta x=0text{ m}$$
Just remember that the average velocity is defined as displacement over time. When an astronaut completes an orbit around the Earth, they make a round-trip, meaning that their displacement (the net change in position) is zero. Your displacement, however, is not zero because you travelled from home to school, so there is a net change in your position.
The reason why is because velocity is a vector quantity, meaning that it has magnitude and direction associated with it. The magnitude of velocity is speed, which is a scalar quantity defined as always being positive, and direction is what gives velocity a positive or negative sign. Therefore, if two different objects have the same speed but different velocities, they are moving in opposite directions.
text{color{#4257b2} boxed{bf Yes}, for two different objects there is a possibility to have the same speed but different velocities if they are traveling in different directions.}
$$
text{color{#4257b2} boxed{bf No}, the statement is not true that if two objects have same velocity they may have different speeds as the velocity of an object is product of speed multiplied by a directional vector that gives the speed a direction, hence in case two objects have same velocity, they should have same speed and same direction of movement.}
$$
Average speed is defined as the distance covered over the elapsed time. Driving for $10 text{km}$ at $15 frac{text{m}}{text{s}}$ takes more time than driving the equal distance at $25 frac{text{m}}{text{s}}$. $textbf{This means that you spend more time travelling at the lower speed. Therefore, the average speed for the entire trip is less than $20 frac{text{m}}{text{s}}$.}$
$textbf{(b)}$ The best explanation is $textbf{A.}$
Since the average speed is averaged over time, the average speed will $text{color{#4257b2} boxed{bf be, less, than} 20 m/s as you will spend a longer time driving at the lower speed. The more weight you give to lower speed the lower will be the average speed. You will cover the 10 km distance in less time at the higher speed than you did at the lower speed.}$
$$
text{color{#4257b2} boxed{bf Optopm,A} is the best answer as more time is spent driving at 15 m/s than at 25 m/s for the same distance, and it will take a longer time at the slower speed to cover the same distance. Statement B is true and irrelevant, and statement C stands false}
$$
In this case, you spend an equal amount of time driving at both speeds. As a result, $textbf{the average speed for the entire trip is exactly $20 frac{text{m}}{text{s}}$,}$ which is also the average of $15 frac{text{m}}{text{s}}$ and $25 frac{text{m}}{text{s}}$.
$textbf{(b)}$ The best explanation is $textbf{C.}$
Since you will spend equal amount of time at both low and high speed so they will nullify the effect and the average speed will be$text{color{#4257b2} boxed{bf equal,to} 20 m/s. The average speed is therefore the mean value of the two speeds.}$
C is the best answer i.e. $text{color{#4257b2} boxed{bf equal, time, is, spent, at, 15 m/s, and, 25, m/s} because that fact is stated in the question. Statements A and B are both false.}$
$$
begin{align*}
v&=dfrac{d}{t}\
&=dfrac{100 text{m}}{53.12 text{s}}\
&=1.88253012 frac{text{m}}{text{s}}\
&=quadboxed{2 frac{text{m}}{text{s}}}\
end{align*}
$$
In miles per hour, her speed was
$$
begin{align*}
2 frac{text{m}}{text{s}}&=2 frac{text{m}}{text{s}}left(dfrac{3600 text{s}}{1 text{h}}right)left(dfrac{1 text{km}}{1000 text{m}}right)left(dfrac{1 text{mi}}{1.61 text{km}}right)\
&=4.47205 frac{text{mi}}{text{h}}\
&=quadboxed{4 frac{text{mi}}{text{h}}}\
end{align*}
$$
The results were given based on $textbf{the rule for multiplication and division for significant figures.}$
begin{align*}
&circquadtext{In meters per second:} quadboxed{2 frac{text{m}}{text{s}}}\
&circquadtext{In miles per hour:} quadboxed{4 frac{text{mi}}{text{h}}}\
end{align*}
$$
$$
begin{align*}
v&=dfrac{d}{t}=dfrac{2600 text{km}}{300 text{days}}\
&=left(dfrac{2600 text{km}}{300 text{days}}right)timesleft(dfrac{1 text{m}}{10^{-3} text{km}}right)left(dfrac{1 text{day}}{24 text{h}}right)left(dfrac{1 text{h}}{3600 text{s}}right)\
&=0.100308 frac{text{m}}{text{s}}\
&=quadboxed{0.1 frac{text{m}}{text{s}}}\
end{align*}
$$
begin{align*}
boxed{v=0.1 frac{text{m}}{text{s}}}\
end{align*}
$$
$$
begin{align*}
v&=dfrac{d}{t}\
rightarrow d&=vt\
&=12 frac{text{m}}{text{s}}times 5.5 text{s}\
&=quadboxed{66 text{m}}\
end{align*}
$$
begin{align*}
boxed{d=66 text{m}}\
end{align*}
$$
textbf{Concept:}
$$
The average speed is calculated by dividing the distance by elapsed time. With this fact we solve for the distance traveled given the speed and the time.
$$
textbf{Solution:}
$$
Use formula for average speed to calculate the distance traveled
$$
average,speed=frac{distance}{time}
$$
$$
speedtimes time = distance = bigg( 12m/s times 5.5 sbigg)=color{#4257b2} boxed{bf 66m}
$$
Distance=66m
$$
$$
begin{align*}
v&=dfrac{d}{t}\
rightarrow t&=dfrac{d}{v}\
&=dfrac{0.23 text{m}}{0.76 frac{text{m}}{text{s}}}\
&=0.30206 text{s}\
&=quadboxed{0.30 text{s}}\
end{align*}
$$
The final result was written based on $textbf{the rule for significant figures for multiplication and division.}$
begin{align*}
boxed{t=0.30 text{s}}\
end{align*}
$$
textbf{Concept:}
$$
The average speed is calculated by dividing the distance by elapsed time. With this fact we solve for the distance traveled given the speed and the time.
$$
textbf{Solution:}
$$
Solve the average speed equation for the time we get:
$$
average,speed=frac{distance}{time}
$$
$$
time=frac{distance}{avg.,speed}=frac{0.23m}{0.76m/s}=color{#4257b2} boxed{bf 0.30s}
$$
Time=0.30s
$$
$$
begin{align*}
v_{trip}&=dfrac{d_{tot}}{t_{tot}}\
rightarrow t_{tot}&=dfrac{d_{tot}}{v_{trip}}\
&=dfrac{24.0 text{km}}{22.0 frac{text{km}}{text{h}}}\
&=1.090909 text{h}\
&=1.09 text{h}\
end{align*}
$$
The time you spent jogging is
$$
begin{align*}
t_{jog}&=dfrac{d_{jog}}{v_{jog}}\
&=dfrac{8.00 text{km}}{9.50 frac{text{km}}{text{h}}}\
&=0.842105 text{h}\
&=0.84 text{h}\
end{align*}
$$
$$
begin{align*}
t_{tot}&=t_{jog}+t_{car}\
t_{car}&=t_{tot}-t_{jog}\
&=1.09 text{h}-0.84 text{h}\
&=0.25 text{h}\
end{align*}
$$
Now we can finally calculate the average speed of the car:
$$
begin{align*}
v_{car}&=dfrac{d_{car}}{t_{car}}\
&=dfrac{16.0 text{km}}{0.25 text{h}}\
&=quadboxed{64.0 frac{text{km}}{text{h}}}\
end{align*}
$$
As always, the results were given based on $textbf{the rule for multiplication and division for significant figures.}$
begin{align*}
boxed{v_{car}=64.0 frac{text{km}}{text{h}}}\
end{align*}
$$
$$
begin{align*}
v’&=dfrac{d’}{t}\
rightarrow t&=dfrac{d’}{v’}\
&=dfrac{5.0 text{m}}{1.3 frac{text{m}}{text{s}}}\
&=3.84615 text{s}\
&=3.8 text{s}
end{align*}
$$
Hence, the dog has travelled
$$
begin{align*}
v_{dog}&=dfrac{d}{t}\
rightarrow d&=v_{dog}t\
&=3.0 frac{text{m}}{text{s}}times 3.8 text{s}\
&=11.4 text{m}\
&=quadboxed{11 text{m}}\
end{align*}
$$
The results were given based on $textbf{the rule for multiplication and division for significant figures.}$
begin{align*}
boxed{d=11 text{m}}\
end{align*}
$$
textbf{Concept:}
$$
Initially calculate the time that has passed before the owners meet each other later determine the distance the dog will cover if it continues running at constant speed over that time interval.
$$
textbf{Solution:}
$$
Calculate the time required by each to walk 5m before they meet each other
$$
elapsed,time=frac{distance}{average,speed}=frac{5m}{1.3m/s}=color{#4257b2} boxed{bf 3.8s}
$$
Now calculate the distance the dog run
$$
d=v_{av}Delta t =(3.0m/s)(3.8s)=color{#4257b2} boxed{bf 11m}
$$
d=11m
$$
This is an example of when the average speed of the entire trip is equal to the mean value of speeds. However, that’s the case only when an object moves at different speeds for an equal amount of time. $textbf{The key here is to know and understand the definition of average speed: it’s speed averaged over a given period of time, not the average of speeds.}$
$$
begin{align*}
circquad d_{1}&=v_{1}t\
&=20.0 frac{text{m}}{text{s}}times 6.00cdot 10^{2} text{s}\
&=12 000 text{m}\
&=12.0cdot 10^{3} text{m}
\
\
\
circquad d_{2}&=v_{2}t\
&=30.0 frac{text{m}}{text{s}}times 6.00cdot 10^{2} text{s}\
&=18 000 text{m}\
&=18.0cdot 10^{3} text{m}\
\
\
\
rightarrow d_{tot}&=d_{1}+d_{2}\
&=30.0cdot 10^{3} text{m}
end{align*}
$$
Therefore, the average speed is
$$
begin{align*}
v&=dfrac{d_{tot}}{t_{tot}}\
&=dfrac{30.0cdot 10^{3} text{m}}{1.20cdot 10^{3} text{s}}\
&=quadboxed{25.0 frac{text{m}}{text{s}}}\
end{align*}
$$
$textbf{(b)}$ $boxed{v=dfrac{d_{tot}}{t_{tot}}=25.0 frac{text{m}}{text{s}}}$
Since the time intervals remains same, equal times is spend at 20 m/s and 30 m/s, and your average speed will be $text{color{#4257b2} boxed{bf equal ;to; 25.0 m/s}.}$
Find the final velocity by dividing total distance by time passes
$$
textbf{Solution:}
$$
$$
v_{av}=frac{v_1Delta T_1+ v_1Delta T_1}{Delta T_1+Delta T_2}
$$
$$
=frac{(20m/s)(10min times 60s/min) + (30m/s)(600s)}{600+600s}=color{#4257b2} boxed{bf 25m/s}
$$
v_{av}=25m/s
$$
Remember that the definition of average speed is speed averaged over a period of time, which is not the same as the average of speeds.
$$
begin{align*}
circquad t_{1}&=dfrac{d}{v_{1}}\
&=dfrac{1}{20.0 frac{text{m}}{text{s}}}times 16.1cdot 10^{3} text{m}\
&=8.05cdot 10^{2} text{s}\
\
\
\
circquad t_{2}&=dfrac{d}{v_{2}}\
&=dfrac{1}{30.0 frac{text{m}}{text{s}}}times 16.1cdot 10^{3} text{m}\
&=536.6666 text{s}\
&=5.37cdot 10^{2} text{s}\
\
\
\
rightarrow t_{tot}&=t_{1}+t_{2}\
&=13.42cdot 10^{2} text{s}\
end{align*}
$$
Thus, the average speed is
$$
begin{align*}
v&=dfrac{d_{tot}}{t_{tot}}\
&=dfrac{32.2cdot 10^{3} text{m}}{13.42cdot 10^{2} text{s}}\
&=23.99404 frac{text{m}}{text{s}}\
&=quadboxed{24.0 frac{text{m}}{text{s}}}\
end{align*}
$$
$textbf{(b)}$ $boxed{v=dfrac{d_{tot}}{t_{tot}}=24.0 frac{text{m}}{text{s}}}$
textbf{Concept:}
$$
Calculate the average speed by calculating the total distance traveled and dividing it by the total time elapsed.
$$
textbf{Solution:}
$$
The time intervals are different but distance intervals are the same. More time is spent at lower speed then higher. Because the average speed is a time weighted average, it will be $text{color{#4257b2} boxed{bf less, than,25.0, m/s}.}$
textbf{Concept:}
$$
By dividing the total distance by time we get
$$
textbf{Solution:}
$$
$$
v_{av}=frac{d_1+d_2}{Delta t_1+Delta t_2}=frac{ d_1+d_2}{(frac{d_1}{s_1})+(frac{d_2}{s_2})}=frac{20km}{frac{10km}{20m/s}+frac{10km}{30m/s}}=color{#4257b2} boxed{bf 24m/s}
$$
v_{av}=24m/s
$$
A horizontal line represents a situation when an object doesn’t change position over time; such a line has zero slope, which corresponds to zero average velocity.
A vertical line would represent a change in position without a change in time. That scenario isn’t possible because it would require an object moving at infinite average velocity; such an object doesn’t exist.
$textbf{(b)}$ No; a vertical line would correspond to an object moving at infinite average velocity.
In case $textbf{B}$, the graph for Train 1 has a negative slope, which means that it has a negative velocity, and its position decreases with time ($x_{i}>x_{f}$), whereas the graph for Train 2 has a positive slope, which corresponds to positive velocity, and its position increases with time ($x_{i}<x_{f}$).
In case $textbf{C}$ we have the same situation, only this time Train 1 has a positive velocity, and Train 2 has a negative velocity.
In case $textbf{A}$, Train 1 travels at a greater velocity than Train 2. During the same time interval, Train 1 covers significantly more distance than Train 2, so as time progresses, the two trains will get farther apart.
In case $textbf{C}$, the trains move in different directions, so it’s clear that they get farther apart as they travel.
The only case in which the graphs intersect is $textbf{B}$; in that case the trains collide.
$textbf{(b)}$ Cases $textbf{A}$ and $textbf{C}$.
$textbf{(c)}$ Case $textbf{B}$.
When the two bodies move in opposite direction then their position time graph is has one positive and the other negative slope. In such a case one trains travel with opposite directions has a positive velocity and the other has a negative velocity. We conclude the trains travel in opposite directions in cases $text{color{#4257b2} boxed{bf B, and, C}.}$
$circ$ In case $textbf{A}$, both trains have a positive velocity.
$circ$ In case $textbf{B}$, Train 2 has a positive velocity.
$circ$ In case $textbf{C}$, Train 1 has a positive velocity.
$circ$ In case $textbf{A}$, none of the trains have a negative velocity.
$circ$ In case $textbf{B}$, Train 1 has a negative velocity.
$circ$ In case $textbf{C}$, Train 2 has a negative velocity.
$textbf{(b)}$ $textbf{A}$: none; $textbf{B}$: Train 1; $textbf{C}$: Train 2.
$textbf{(b)}$ Zero velocity.
$textbf{(c)}$ Positive velocity.
$textbf{(d)}$ Negative velocity.
Segment A has $text{color{#4257b2} boxed{bf positive} velocity as the slope for this segment is positive.}$
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}\
&=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{2 text{m}-0 text{m}}{1 text{s}-0 text{s}}\
&=quadboxed{2 frac{text{m}}{text{s}}}\
end{align*}
$$
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{2 text{m}-2 text{m}}{2 text{s}-1 text{s}}\
&=quadboxed{0 frac{text{m}}{text{s}}}\
end{align*}
$$
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{3 text{m}-2 text{m}}{3 text{s}-2 text{s}}\
&=quadboxed{1 frac{text{m}}{text{s}}}\
end{align*}
$$
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{0 text{m}-3 text{m}}{5 text{s}-3 text{s}}\
&=quadboxed{-1.5 frac{text{m}}{text{s}}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{v_{avg}=2 frac{text{m}}{text{s}}}\
textbf{(b)} quad &boxed{v_{avg}=0 frac{text{m}}{text{s}}}\
textbf{(c)} quad &boxed{v_{avg}=1 frac{text{m}}{text{s}}}\
textbf{(d)} quad &boxed{v_{avg}=-1.5 frac{text{m}}{text{s}}}\
end{align*}
$$
textbf{Concept:}
$$
Calculate the velocity by calculating the slopes from the graph at each mentioned points
$$
textbf{Solution:}
$$
For slope at point A
$$
v_{av}=frac{Delta x}{Delta t}=frac{2m}{1s}=color{#4257b2} boxed{bf 2.0m/s}
$$
Similarly calculate for rest of the point B,C and D we get
For point B
$$
v_{av}=frac{Delta x}{Delta t}=frac{0.0m}{1s}=color{#4257b2} boxed{bf 0.0m/s}
$$
For point C
$$
v_{av}=frac{Delta x}{Delta t}=frac{1m}{1s}=color{#4257b2} boxed{bf 1m/s}
$$
For point D
$$
v_{av}=frac{Delta x}{Delta t}=frac{-3.0m}{2s}=color{#4257b2} boxed{bf -1.5m/s}
$$
v_{av,A}=2.0m/s,,v_{av,B}=0.0m/s,,v_{av,C}=1.0m/s,,v_{av,D}=-1.5m/s
$$
$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$
In our case, the initial position is $x_{i}=5.0 text{m}$ and the velocity is $v=3.5 frac{text{m}}{text{s}}$. Therefore, the equation of motion is
$$
begin{align*}
x_{f}=5.0 text{m}+left(3.5 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
begin{align*}
begin{tabular}{|c|c|}
hline
textbf{Time (s)} & textbf{Position (m)}\
hline
0 & 5\
hline
2 & 12\
hline
4 & 19\
hline
6 & 26\
hline
end{tabular}
end{align*}
$quadquadquadquadquadquadquad boxed{x_{f}=5.0 text{m}+left(3.5 frac{text{m}}{text{s}}right)times t}$
$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$
In our case, the initial position is $x_{i}=3.1 text{m}$ and the velocity is $v=-2.7 frac{text{m}}{text{s}}$. Therefore, the equation of motion is
$$
begin{align*}
x_{f}=3.1 text{m}+left(-2.7 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
Note that the velocity is negative, which means that $textbf{the slope of the position-time graph will be negative.}$
begin{align*}
begin{tabular}{|c|c|}
hline
textbf{Time (s)} & textbf{Position (m)}\
hline
0 & 3.1\
hline
1 & 0.4\
hline
2 & -2.3\
hline
3 & -5\
hline
4 & -7.7\
hline
5 & -10.4\
hline
6 & -13.1\
hline
end{tabular}
end{align*}
$quadquadquadquadquadquadquad boxed{x_{f}=3.1 text{m}+left(-2.7 frac{text{m}}{text{s}}right)times t}$
$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$
To plot the motion of the object in question, we will divide the problem into three parts.
begin{align*}
x_{f}^{(1)}=1.5 text{m}+left(2.2 frac{text{m}}{text{s}}right)times t\
end{align*}
The corresponding two positions for the values of $t=0$ and $t=1 text{s}$ are given in the following table.\
begin{align*}
begin{tabular}{|c|c|}
hline
textbf{Time (s)} & textbf{Position (m)}\
hline
0 & 1.5\
hline
1 & 3.7\
hline
end{tabular}
end{align*}
\
Because the velocity is positive, textbf{this part of motion will have a positive slope on a position-time graph.}\
$$
begin{align*}
x_{f}^{(2)}=3.7 text{m}=const.\
end{align*}
$$
$textbf{This part of motion is represented by a horizontal line because there is no change in position over time.}$
begin{align*}
x_{f}^{(3)}=3.7 text{m}+left(-3.7 frac{text{m}}{text{s}}right)times t\
end{align*}
textbf{However, be careful when choosing the values for $t$.} It would be a mistake to plug in the values $t=2 text{s}, 3 text{s}, 4 text{s}, 5 text{s}$ into the equation of motion. Instead, think of this as a completely new motion from $t=0$ to $t=3 text{s}$, with the initial position being $x_{i}=3.7 text{m}$. This will result in correct values for the final positions, but when plotting the graph (or rather, adding this part of motion to the already existing graph), textbf{start at $t=2 text{s}$ and finish at $t=5 text{s}.$} If this is confusing, it will make more sense once you see the position-time graph.\
Therefore, the values for $x_{f}^{(3)}$ for this part of motion are:\
begin{align*}
begin{tabular}{|c|c|}
hline
textbf{Time (s)} & textbf{Position (m)}\
hline
0 & 3.7\
hline
1 & 0\
hline
2 & -3.7\
hline
3 & -7.4\
hline
end{tabular}
end{align*}
$quadquadquadquadquadquadquadquadquadquad boxed{x_{f}^{(i)}=x_{i}^{(i)}+v^{(i)}t} quad , quad (i)=1,2,3$
Click here to see the position-time graph.
$$
begin{align*}
boxed{x_{i,A}=35 text{m}}\
end{align*}
$$
$$
begin{align*}
boxed{x_{i,B}=10 text{m}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{x_{i,A}=35 text{m}}\
textbf{(b)} quad &boxed{x_{i,B}=10 text{m}}\
end{align*}
$$
From graph we found that position of object B. By taking the values of y-intercept of the position-time graph for object A we see that it has a value of $text{color{#4257b2} boxed{bf 35 m} at t = 0.}$
From graph we found that position of object B. By taking the values of the y-intercept of the position-time graph for object B we see that it has a value of $text{color{#4257b2} boxed{bf 10 m} at t = 0.}$
As we already know, $textbf{the slope of a straight line on a position-time graph corresponds to the average velocity during a chosen time period.}$ The slope of a line is equal to its rise over run, i.e., the change in vertical axis value over the change in horizontal axis value.
$$
begin{align*}
text{slope}=dfrac{text{rise}}{text{run}} quad Longrightarrowquad v_{avg}=dfrac{Delta x}{Delta t}\
end{align*}
$$
Because the slope is the same anywhere along a straight line, we can chose any pair of points to calculate its value.
$$
begin{align*}
v_{A}&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{5 text{m}-35 text{m}}{3.5 text{s}}\
&=quadboxed{-8.6 frac{text{m}}{text{s}}}\
end{align*}
$$
$$
begin{align*}
v_{B}&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{25 text{m}-10 text{m}}{3.5 text{s}}\
&=quadboxed{4.3 frac{text{m}}{text{s}}}\
end{align*}
$$
begin{align*}
&textbf{(a)} quadboxed{v_{A}=-8.6 frac{text{m}}{text{s}}}\
\
&textbf{(b)} quadboxed{v_{B}=4.3 frac{text{m}}{text{s}}}\
end{align*}
$$
The initial pair of points is $(t_{i}, x_{i})=(1 text{s}, 2 text{m})$ and the final pair of points is $(t_{f}, x_{f})=(3 text{s}, 3 text{m})$, which means that the father’s average velocity is
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}\
&=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{3 text{m}-2 text{m}}{3 text{s}-1 text{s}}\
&=dfrac{1 text{m}}{2 text{s}}\
&=quadboxed{0.5 frac{text{m}}{text{s}}}\
end{align*}
$$
begin{align*}
boxed{v_{avg}=0.5 frac{text{m}}{text{s}}}\
end{align*}
$$
textbf{Concept:}
$$
Calculate the velocity by dividing the displacement by the elapsed time. For the instance where $t = 1 s$ the position of the father is 2 m, and for $t=3s$ his position is 3 m.
$$
textbf{Solution:}
$$
Obtain velocity by dividing displacement with time
$$
v_{av}=frac{Delta x}{Delta t}=frac{3-2m}{3-1s}=color{#4257b2} boxed{bf 0.5m/s}
$$
v_{av}=0.5m/s
$$
Imagine, for instance, that you and your friend are standing side-by-side facing opposite directions and start walking. One of you will have a negative velocity, whereas the other will have a positive velocity, but both of you will have the same initial position.
text{color{#4257b2} boxed{bf Yes}, two objects can have two different velocities while having same starting position as an object’s velocity is completely distinct from its initial position.}
$$
Again, there is no reason why the two objects in question couldn’t have the same velocity based on their initial positions; the initial position has nothing to do with velocity; it’s merely the position at $t=0$ and nothing more. The only two things that matter are that both objects have the same speed and travel in the same direction.
text{color{#4257b2} boxed{bf Yes}, two objects can have two different velocities while having same starting position as an object’s velocity is completely distinct from its initial position. For the objects to have the same velocity only requirement is that they move with same speed and be traveling in the same direction.}
$$
The slope of the position-time graph stays the same because it represents the velocity of an object. If the velocity is constant, i.e., doesn’t change, there is no change in slope.
You can easily verify this by plotting a position-time graph for motion with constant velocity, and then shifting the whole line up; this results in two parallel lines.
This is also easily verified by plotting the position-time graph for constant velocity; the slope gets steeper at the point of your choosing, but the initial position doesn’t change.
$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$
which means that the first fish has the velocity of $v_{1}=-1.2 frac{text{m}}{text{s}}$ and the velocity of the second fish is $v_{2}=-2.7 frac{text{m}}{text{s}}$.
Although the velocity of the second fish is more negative than the velocity of the first fish, $textbf{what matters here is the absolute value of their velocities, or, in other words, their speed.}$
This means that the first fish has the speed of $1.2 frac{text{m}}{text{s}}$ and the second fish has the speed of $2.7 frac{text{m}}{text{s}}$. We can, therefore, conclude that $textbf{the second fish is moving faster.}$
$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$
we can easily see that the velocity of the first person is $v_{1}=-1.1 frac{text{m}}{text{s}}$, and the velocity of the second person is $v_{2}=1.7 frac{text{m}}{text{s}}$. $textbf{To determine which of these two people is moving faster, we’ll take the absolute value (or the magnitude) of their velocities.}$
The magnitudes of these two velocities are $1.1 frac{text{m}}{text{s}}$ for the first person, and $1.7 frac{text{m}}{text{s}}$, for the second person. $textbf{Therefore, we can conclude that the second person is moving faster.}$
$textbf{(b)}$ The first person will be at $x=0$ at some point in the future.
$$
begin{align*}
y=b+axquadLongrightarrowquad x_{f}=x_{i}+vt\
end{align*}
$$
by reading the data off the position-time graph.
$$
begin{align*}
boxed{x_{i}=35 text{m}}\
end{align*}
$$
The slope of a straight line is its rise over its run, which corresponds to $textbf{the change in position over the elapsed time}$ on the position-time graph. We can choose any two pairs of points $(t,x)$ on the graph, but the easiest are $(t_{1}=0 text{s}, x_{1}=35 text{m})$ and $(t_{2}=3.5 text{s}, x_{2}=5 text{m})$. Therefore, the average velocity is
$$
begin{align*}
v&=dfrac{Delta x}{Delta t}=dfrac{x_{2}-x_{1}}{t_{2}-t_{1}}\
&=dfrac{5 text{m}-35 text{m}}{3.5 text{s}}\
&=quad boxed{-8.6 frac{text{m}}{text{s}}}\
end{align*}
$$
$$
begin{align*}
boxed{x_{f}=35 text{m}-left(8.6 frac{text{m}}{text{s}}right)times t}\
end{align*}
$$
begin{align*}
boxed{x_{f}=35 text{m}-left(8.6 frac{text{m}}{text{s}}right)times t}\
end{align*}
$$
textbf{Concept:}
$$
We determine the velocity of an object from the slope of its position-time graph. The position-time graph for object A shows that it has a negative slope hence negative velocity. To calculate the velocity, choose two points on the line and find the slope by dividing the rise by the run.
$$
textbf{Solution:}
$$
Select two points on the graph say (0s, 35m) and (3.5s, 5m) and find compute the velocity
$$
v_{av,A}=frac{rise}{run}=frac{5-35m}{3.5-0s}=color{#4257b2} boxed{bf -8.6m/s}
$$
Now use the starting position of 35m to write the equation of motion
$$
color{#4257b2} boxed{bf x_f=35m-(8.6m/s)t}
$$
x_f=35m-(8.6m/s)t
$$
$$
begin{align*}
y=b+axquadLongrightarrowquad x_{f}=x_{i}+vt\
end{align*}
$$
by reading the data off the position-time graph.
$$
begin{align*}
boxed{x_{i}=10 text{m}}\
end{align*}
$$
The slope of a straight line is its rise over its run, which corresponds to $textbf{the change in position over the elapsed time}$ on the position-time graph. We can choose any two pairs of points $(t,x)$ on the graph, but the easiest are $(t_{1}=0 text{s}, x_{1}=10 text{m})$ and $(t_{2}=3.5 text{s}, x_{2}=25 text{m})$. Therefore, the average velocity is
$$
begin{align*}
v&=dfrac{Delta x}{Delta t}=dfrac{x_{2}-x_{1}}{t_{2}-t_{1}}\
&=dfrac{25 text{m}-10 text{m}}{3.5 text{s}}\
&=quad boxed{4.3 frac{text{m}}{text{s}}}\
end{align*}
$$
$$
begin{align*}
boxed{x_{f}=10 text{m}+left(4.3 frac{text{m}}{text{s}}right)times t}\
end{align*}
$$
begin{align*}
boxed{x_{f}=10 text{m}+left(4.3 frac{text{m}}{text{s}}right)times t}\
end{align*}
$$
textbf{Concept:}
$$
We find the velocity of an object from the slope of its position-time graph. Object B’s position on the position-time graph shows a positive slope hence it has positive velocity. To calculate the velocity, choose two points on the line and calculate the slope by dividing the rise by the run.
$$
textbf{Solution:}
$$
Select two points on the graph say (0s, 10m) and (3.5s, 25m) and find compute the velocity
$$
v_{av,B}=frac{rise}{run}=frac{25-10m}{3.5-0s}=color{#4257b2} boxed{bf 4.3m/s}
$$
Now use the starting position of 10m to write the equation of motion
$$
color{#4257b2} boxed{bf x_f=10m+(4.3m/s)t}
$$
x_f=10m+(4.3m/s)t
$$
$$
begin{align*}
y=& b+ax\
&Downarrow\
x_{f}=& x_{i}+vt\
end{align*}
$$
the rabbit’s equation of motion is
$$
begin{align*}
boxed{x_{f}=8.1 text{m}-left(1.6 frac{text{m}}{text{s}}right)times t}\
end{align*}
$$
begin{align*}
boxed{x_{f}=8.1 text{m}-left(1.6 frac{text{m}}{text{s}}right)times t}\
end{align*}
$$
textbf{Concept:}
$$
In equation of motion, time multiplied by velocity then the initial position is added to the equation to get the final position. Use the given initial position and velocity to develop the equation of motion for the rabbit.
$$
textbf{Solution:}
$$
With initial position of 8.1m and velocity of -1.6m/s when we substitute values in the equation we get
$$
color{#4257b2} boxed{bf x_f=8.1m-(1.6m/s)t}
$$
x_f=8.1m-(1.6m/s)t
$$
$$
begin{align*}
x=6.0 text{m}+left(4.5 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
Therefore, at $t=2.0 text{s}$, the corresponding position is
$$
begin{align*}
x&=6.0 text{m}+left(4.5 frac{text{m}}{text{s}}right)times 2.0 text{s}\
&=quadboxed{15 text{m}}\
end{align*}
$$
$$
begin{align*}
24 text{m}=6.0 text{m}+left(4.5 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
Rearranging the terms and solving for $t$, we obtain the following result:
$$
begin{align*}
t&=dfrac{24 text{m}-6.0 text{m}}{4.5 frac{text{m}}{text{s}}}\
&=dfrac{18}{4.5} text{s}\
&=quadboxed{4.0 text{s}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{x=15 text{m}}\
\
textbf{(b)} quad &boxed{t=4.0 text{s}}\
end{align*}
$$
textbf{Concept:}
$$
We multiply time with velocity and then add initial position to get the final position With given data that the bicycle starts at 6.0 m at $t=0s$ and travels with a constant velocity of 4.5 m/s we can use equation of motion and substitute values to answer the questions.
$$
textbf{Solution:}
$$
$$
x_f=(6.0)+(4.5m/s)(2.0s)=color{#4257b2} boxed{bf 15.0m}
$$
textbf{Concept:}
$$
Solve for $t$ the equation of motion to find it given $x_f=24m$
$$
textbf{Solution:}
$$
$$
x_f=x_i+vt
$$
$$
x_f-x_i=vt Rightarrow t=frac{ x_f-x_i}{v}=frac{24.0-6.0m}{4.5m/s}=color{#4257b2} boxed{bf 4.0s}
$$
t=4.0s
$$
$$
begin{align*}
x=-9.2 text{m}+left(1.5 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
Therefore, at $t=3.5 text{s}$, the corresponding position is
$$
begin{align*}
x&=-9.2 text{m}+left(1.5 frac{text{m}}{text{s}}right)times 3.5 text{s}\
&=-3.95 text{m}\
&=quadboxed{-4.0 text{m}}\
end{align*}
$$
Note that we applied $textbf{the rule for addition and subtraction for significant figures}$ when writing the final result.
$$
begin{align*}
0 text{m}=-9.2 text{m}+left(1.5 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
Rearranging the terms and solving for $t$, we obtain the following result:
$$
begin{align*}
t&=dfrac{9.2 text{m}}{1.5 frac{text{m}}{text{s}}}\
&=6.133333 text{s}\
&=quadboxed{6.1 text{s}}\
end{align*}
$$
In this case, we applied $textbf{the rule for multiplication and division for significant figures}$ when writing the final result.
begin{align*}
textbf{(a)} quad &boxed{x=-4.0 text{m}}\
\
textbf{(b)} quad &boxed{t=6.1 text{s}}\
end{align*}
$$
textbf{Concept:}
$$
We multiply time with velocity and then add initial position to get the final position With given data that the float starts at -9.2m at $t=0s$ and travels with a constant velocity of 1.5 m/s we can use equation of motion and substitute values to answer the questions.
$$
textbf{Solution:}
$$
$$
x_f=(-9.2)+(1.5m/s)(3.5s)=color{#4257b2} boxed{bf -4.0m}
$$
textbf{Concept:}
$$
Solve for $t$ the equation of motion to find it given $x_f=0m$
$$
textbf{Solution:}
$$
$$
x_f=x_i+vt
$$
$$
x_f-x_i=vt Rightarrow t=frac{ x_f-x_i}{v}=frac{0-(-9.2)m}{1.5m/s}=color{#4257b2} boxed{bf 6.1s}
$$
t=6.1s
$$
So, knowing that Cleo’s equation of motion is given by
$$
begin{align*}
x=-12.1 text{m}+left(5.2 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
we can easily find his position at $t=1.6 text{s}$ by plugging that value into the equation of motion:
$$
begin{align*}
x&=-12.1 text{m}+left(5.2 frac{text{m}}{text{s}}right)times 1.6 text{s}\
&=-12.1 text{m}+8.32 text{m}\
&=-3.78 text{m}\
&=quadboxed{-3.8 text{m}}\
end{align*}
$$
When writing the final result, we used $textbf{the rule for addition and subtraction for significant figures.}$
$$
begin{align*}
3.0 text{m}=-12.1 text{m}+left(5.2 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
Solving the equation for $t$ leads to the moment when Cleo reaches the stick:
$$
begin{align*}
t&=dfrac{3.0 text{m}+12.1 text{m}}{5.2 frac{text{m}}{text{s}}}\
&=dfrac{15.1}{5.2} text{s}\
&=2.903846 text{s}\
&=quadboxed{2.9 text{s}}\
end{align*}
$$
In this case, we used $textbf{the rule for multiplication and division for significant figures}$ when writing the final result.
begin{align*}
textbf{(a)} quad &boxed{x=-3.8 text{m}}\
\
textbf{(b)} quad &boxed{t=2.9 text{s}}\
end{align*}
$$
textbf{Concept:}
$$
We multiply time with velocity and then add initial position to get the final position With given data that the Cleo’s starts at -12.1m at $t=1.6s$ and travels with a constant velocity of 5.2 m/s we can use equation of motion and substitute values to answer the questions.
$$
textbf{Solution:}
$$
$$
x_f=(-12.1)+(5.2m/s)(1.6s)=color{#4257b2} boxed{bf -3.8m}
$$
textbf{Concept:}
$$
Solve for $t$ the equation of motion to find it given $x_f=3m$
$$
textbf{Solution:}
$$
$$
x_f=x_i+vt
$$
$$
x_f-x_i=vt Rightarrow t=frac{ x_f-x_i}{v}=frac{0-(12.1)m}{5.2m/s}=color{#4257b2} boxed{bf 2.9s}
$$
t=2.9s
$$
$$
begin{align*}
y=b+ax quadLongrightarrowquad x_{f}=x_{i}+vt\
end{align*}
$$
The first player moves at a velocity of $v_{1}=-3.1 frac{text{m}}{text{s}}$, and the second player has a velocity of $v_{2}=2.8 frac{text{m}}{text{s}}$.
$textbf{Regardless of the fact that the first player has a negative velocity, he is moving faster than the second player.}$ The reason for this is that when it comes to determining who is faster, we are interested in just the magnitude (the absolute value) of velocity, which is speed, and not the direction of motion.
$$
begin{align*}
x_{1}=x_{2}\
end{align*}
$$
Rearranging the terms and solving for $t$, we obtain the result:
$$
begin{align*}
0.1 text{m} -left(3.1 frac{text{m}}{text{s}}right)times t&=-6.3 text{m}+left(2.8 frac{text{m}}{text{s}}right)times t\
left(2.8 frac{text{m}}{text{s}}+3.1 frac{text{m}}{text{s}}right)times t&=0.1 text{m}+6.3 text{m}\
rightarrowquad t&=dfrac{6.4 text{m}}{5.9 frac{text{m}}{text{s}}}\
&=1.08475 text{s}\
&=quadboxed{1.1 text{s}}\
end{align*}
$$
When writing the final result, we used $textbf{the rule for multiplication and division for significant figures.}$
begin{align*}
textbf{(a)} quad &text{The first player is moving faster.}\
textbf{(b)} quad &boxed{t=1.1 text{s}}\
end{align*}
$$
textbf{Concept:}
$$
We need to draw the graphs for the two football players with the given equations of motion. The graphs are not mandatory to answer the questions but will be helpful in picturing the situation. Lastly, use the equations of motion to calculate which player is moving faster and calculate the time at which the two football players will collide. The two equations are:
$$
textbf{Solution:}
$$
$$
x_{f,1}=x_{i,1}+v_1 t=(0.1m)+(-3.1m/s)t
$$
$$
x_{f,2}=x_{i,2}+v_2 t=(0.1m)+(-3.1m/s)t
$$
In the equations of motion we multiply velocity with time and add initial position of the object to find the final position. From the graph we can see that the velocity of the first football player is$-3.1m/s$ and the velocity of the second football player is $2.8m/s$. The speed of each player is the absolute amount of his The speed of each player is the absolute amount of his velocity, so we find that football $text{color{#4257b2} boxed{bf player 1} is traveling faster than football player 2.}$
textbf{Concept:}
$$
Let the final position of the players be equal to each other then
$$
textbf{Solution:}
$$
$$
x_{f,1}=x_{f,2}
$$
$$
x_{i,1}+v_1 t=x_{i,2}+ v_2 t
$$
Simplifying the equation we get
$$
-v_2t+v_1t= x_{i,2}- x_{i,1}
$$
$$
(v_1-v_2)t= x_{i,2}- x_{i,1}
$$
$$
t=frac{ x_{i,2}- x_{i,1}}{(v_1-v_2)}=frac{-6.3-(0.1)m}{-3.1-(2./m/s)}=color{#4257b2} boxed{bf 1.1s}
$$
t=1.1s
$$
$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$
Knowing the equations of motion for both players, it’s easy to see that their positions at $t=0$ are
$$
begin{align*}
x_{1} (t=0)&=-8.2 text{m}\
x_{2} (t=0)&=-7.3 text{m}\
end{align*}
$$
where $x_{1}$ is the position of the first player, and $x_{2}$ is the position of the second player.
The first player is further away from the chosen origin than the second player, which means that he has to cover more distance to reach the ball which is at $x_{b}=5.0 text{m}$. Specifically, the respective distances of the first and second player from the ball are
$$
begin{align*}
d_{1}&=8.2 text{m}+5.0 text{m}=13.2 text{m}\
d_{2}&=7.3 text{m}+5.0 text{m}=12.3 text{m}\
end{align*}
$$
Hence, we can conclude that $textbf{the second player is closer to the ball at $t=0$.}$
$$
begin{align*}
x_{1}=x_{2}\
end{align*}
$$
Plugging in the values and rearranging the terms yields
$$
begin{align*}
-8.2 text{m}+left(4.2 frac{text{m}}{text{s}}right)times t&=-7.3 text{m}+left(3.9 frac{text{m}}{text{s}}right)times t\
left(4.2 frac{text{m}}{text{s}}-3.9 frac{text{m}}{text{s}}right)times t&=8.2 text{m}-7.3 text{m}\
rightarrowquad t&=dfrac{0.9 text{m}}{0.3 frac{text{m}}{text{s}}}\
&=quadboxed{3.0 text{s}}\
end{align*}
$$
Using the equation of motion of the first player, we get
$$
begin{align*}
x_{1} (t=3.0 text{s})&=-8.2 text{m}+4.2 frac{text{m}}{text{s}}cdot 3.0 text{s}\
&=-8.2 text{m}+12.6 text{m}\
&=quadboxed{4.4 text{m}}\
end{align*}
$$
Additionally, we can show that the second player has the same position as the first:
$$
begin{align*}
x_{2} (t=3.0 text{s})&=-7.3 text{m}+3.9 frac{text{m}}{text{s}}cdot 3.0 text{s}\
&=-7.3 text{m}+11.7 text{m}\
&=quadboxed{4.4 text{m}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &text{At $t=0$, the second player is closer to the ball.}\
\
textbf{(b)} quad &boxed{t=3.0 text{s}}\
\
textbf{(c)} quad &boxed{x_{1}=x_{2}=4.4 text{m}}\
end{align*}
$$
$$
begin{align*}
v=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
end{align*}
$$
The displacement of the golf cart from $t=0$ to $t=5 text{s}$ is equal to its displacement from $t=5 text{s}$ to $t=10 text{s}$ because the elapsed time is the same in both cases and equals $Delta t=5 text{s}$, and there is no mention of change in velocity.
$$
begin{align*}
v_{avg}=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
end{align*}
$$
Just by looking at the formula we can conclude that $textbf{only the net change in position during the time of motion is important when calculating the average velocity. This means that it’s possible to be at rest at some point during motion and still have an average velocity different than zero,}$ i.e., as long as the net change in position is different than zero, the average velocity will also be different than zero; the magnitude of instantaneous velocity doesn’t matter.
$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$
so it’s easy to see that the velocity of the first dragonfly is $v_{1}=0.75 frac{text{m}}{text{s}}$, whereas the second dragonfly has the velocity of $v_{2}=-1.1 frac{text{m}}{text{s}}$.
$textbf{When determining which dragonfly is faster, we are interested in the magnitude of velocity, which is speed.}$ The speed of the first dragonfly is $0.75 frac{text{m}}{text{s}}$, and the speed of the second dragonfly is $1.1 frac{text{m}}{text{s}}.$ Therefore, despite having a negative velocity, the second dragonfly has greater speed, which means that it’s moving faster than the first one.
$$
begin{align*}
x_{1}&=2.2 text{m}\
x_{2}&=-3.1 text{m}\
end{align*}
$$
and we can immediately conclude that $textbf{the first dragonfly is closer to $x=0$ at $t=0$ than the second one.}$
$textbf{(b)}$ The first dragonfly is closer to $x=0$ at $t=0$ than the second one.
With the given equation of motion graphs can be drawn for the position of dragonflies. The graphs are not msut to have but come handy in solving the problem. Lastly, use the equations of motion to determine which dragonfly is moving faster and find the time at which the two dragonflies
$$x_{f,1}= x_{v,1}+v_1t=(2.2m)+(0.75m/s)t$$
$$x_{f,2}= x_{v,2}+v_1t=(-3.1m)+(-1.10m/s)t$$
textbf{Solution :}\
The above equation adds to the initial position of dragonfly the product of time and its velocity to get the final position. By inspecting equations we can find that the velocity of the first dragonfly is 0.75m/s and the velocity of the second dragonfly is -1.1m/s. Since the negative sing only shows direction of movement, we conclude that color{blue} boxed{bf mbox{dragonfly 2}} is moving faster
With the given equation of motion graphs can be drawn for the position of dragonflies. The graphs are not msut to have but come handy in solving the problem. Lastly, use the equations of motion to determine which dragonfly is moving faster and find the time at which the two dragonflies
$$x_{f,1}= x_{v,1}+v_1t=(2.2m)+(0.75m/s)t$$
$$x_{f,2}= x_{v,2}+v_1t=(-3.1m)+(-1.10m/s)t$$
textbf{Solution :}\
From the equations of motion we know that the number that is added to the right side is the initial position of the dragonfly. By inspection it is clear thatcolor{blue} boxed{bf mbox{dragonfly 1}} is closer to $x=0$ at $t=0$ because 2.2 m is closer to zero than is -3.1m. The two initial positions are represented on the position-time graph with arrows.
Let’s assume that our snail is $5 text{cm}$ long, so during the 10 second interval, it covers $5 text{cm}$. Its average velocity is then
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}\
&=dfrac{0.05 text{m}}{10 text{s}}\
&=quadboxed{5cdot 10^{-3} frac{text{m}}{text{s}}}\
end{align*}
$$
or $0.5 frac{text{cm}}{text{s}}$. This is a somewhat reasonable estimate because research shows that the top speed of a garden snail is $1.3 frac{text{cm}}{text{s}}$
begin{align*}
boxed{v_{avg}=5cdot 10^{-3} frac{text{m}}{text{s}}}\
end{align*}
$$
To make forecast of the speed we will select a time interval and then estimate the displacement during that interval.
$textbf{Solution :}$
If you observe a garden snail crawling for t=10sec, it might travel a distance about equal to its 2-cm body length. Its estimated speed is thus:
$$
v_{av}=frac{Delta x}{Delta t}=frac{0.02m}{10s}=2times 10^{-3}m/scongcolor{#4257b2} boxed{bf 2times 10^{-3}m/s}
$$
2times 10^{-3}m/s
$$
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}\
Delta t&=dfrac{Delta x}{v_{avg}}\
&=dfrac{1 text{m}}{1cdot 10^{2} frac{text{m}}{text{s}}}\
&=quadboxed{0.01 text{s}}\
end{align*}
$$
begin{align*}
boxed{Delta t=0.01 text{s}}\
end{align*}
$$
To get the time elapsed we will take the ratio of the displacement and the average speed. The distance from your finger to your brain assumed to be one meter.
$textbf{Solution :}$
Divide the distance by the average speed:
$$
Delta t=frac{Delta x}{v_{av}}=frac{1m}{1 times 10^2m/s}=color{#4257b2} boxed{bf 0.010s}
$$
Delta t=0.010s
$$
The displacement of object 1 is $2 text{m}$, whereas the displacement of object 2 is $1 text{m}$ Therefore, $textbf{object 1 has the greater displacement than object 2.}$
$^{*}$ Note that the information we are given about the time of each motion is completely unnecessary for determining displacements; by definition, displacement is the net change in position and it has nothing to do with time.
$$
begin{align*}
x_{f}&=x_{i}+vt\
&=7.3 text{m}-1.1 frac{text{m}}{text{s}}cdot 3.5 text{s}\
&=7.3 text{m}-3.85 text{m}\
&=3.45 text{m}\
&=quadboxed{3.5 text{m}}\
end{align*}
$$
Note that we used $textbf{the rule for addition and subtraction for significant figures}$ when writing the final result.
begin{align*}
boxed{x_{f}=3.5 text{m}}\
end{align*}
$$
With equation of motion and given data we can find the final position
$textbf{Solution :}$
From the equation of motion we get:
$$
x_f=x_i+vt=(7.3m)+(-1.1m/s)(3.5s)=color{#4257b2} boxed{bf 3.5m}
$$
$$
x_f= 3.5m
$$
x_f= 3.5m
$$
$$
begin{align*}
v&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{Delta t}\
&=dfrac{62 text{m}-73 text{m}}{12 text{s}}\
&=-0.91666 frac{text{m}}{text{s}}\
&=quadboxed{-0.92 frac{text{m}}{text{s}}}\
end{align*}
$$
Notice that when writing the final result, we used $textbf{the rule for multiplication and division for significant figures.}$
$$
begin{align*}
text{speed}&=abs{v}\
&=abs{-0.92 frac{text{m}}{text{s}}}\
&=quadboxed{0.92 frac{text{m}}{text{s}}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &boxed{v=-0.92 frac{text{m}}{text{s}}}\
\
textbf{(b)} quad &boxed{text{speed}=abs{v}=0.92 frac{text{m}}{text{s}}}\
end{align*}
$$
With the given data we can find velocity by dividing the displacement by the time elapsed.
$textbf{Solution :}$
$$
v_{av}=frac{Delta x}{Delta t}=frac{x_f-x_i}{v_i-v_f}=frac{62-73m}{12-0s}=color{#4257b2} boxed{bf -0.92m/s}
$$
The speed is absolute value of velocity
$textbf{Solution :}$
$$
s=left|-0.92right|=color{#4257b2} boxed{bf 0.92m/s}
$$
s=0.92m/s
$$
begin{align*}
begin{tabular}{|c|c|}
hline
textbf{Time (s)} & textbf{Position (m)}\
hline
0 & 4.3\
hline
1 & 11\
hline
2 & 17.7\
hline
3 & 24.4\
hline
4 & 31.1\
hline
5 & 37.8\
hline
end{tabular}
bigskip\
end{align*}
$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$
where $x_{i}$ is the intercept and $v$ is the slope.
Plugging in the given values, we get the following:
$$
begin{align*}
boxed{x_{f}=4.3 text{m}+left(6.7 frac{text{m}}{text{s}}right)times t}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &text{Click to see the graph.}\
\
textbf{(b)} quad &boxed{x_{f}=4.3 text{m}+left(6.7 frac{text{m}}{text{s}}right)times t}\
end{align*}
$$
begin{align*}
begin{tabular}{|c|c|}
hline
textbf{Time (s)} & textbf{Position (m)}\
hline
0 & 11\
hline
1 & 17.5\
hline
2 & 24\
hline
3 & 30.5\
hline
4 & 37\
hline
5 & 43.5\
hline
end{tabular}
end{align*}
$$
begin{align*}
32 text{m}&=11 text{m}+left(6.5 frac{text{m}}{text{s}}right)times t\
rightarrowquad t&=dfrac{32 text{m}-11 text{m}}{6.5 frac{text{m}}{text{s}}}\
&=3.23077 text{s}\
&=quadboxed{3.2 text{s}}\
end{align*}
$$
When writing the final result, we applied $textbf{the rule for multiplication and division for significant figures.}$
begin{align*}
textbf{(a)} quad &text{Click to see the graph.}\
\
textbf{(b)} quad &boxed{t=3.2 text{s}}\
end{align*}
$$
After going halfway around the track, the girl and her pony have covered half the circumference of the circular track
$$
begin{align*}
d&=dfrac{2rpi}{2}\
&=4.5 text{m}pi\
&=14.137167 text{m}\
&=quadboxed{14 text{m}}\
end{align*}
$$
Their displacement, however, is equal to the diameter of the track
$$
begin{align*}
Delta x&=2r\
&=2times 4.5 text{m}\
&=quadboxed{9.0 text{m}}\
end{align*}
$$
When writing the final results for both the distance covered and the displacement, we used $textbf{the rule for multiplication and division for significant figures.}$
$$
begin{align*}
d&=2rpi\
&=2pitimes 4.5 text{m}\
&=28.27433 text{m}\
&=quadboxed{28 text{m}}\
end{align*}
$$
and the displacement is zero because there is no net change in position; the final and initial point are the same
$$
begin{align*}
boxed{Delta x=0}\
end{align*}
$$
Notice that we again used $textbf{the rule for multiplication and division for significant figures}$ when writing the final result for displacement.
begin{align*}
textbf{(a)} quad &boxed{d=14 text{m}}\
&boxed{Delta x=9.0 text{m}}\
\
textbf{(b)} quad &text{When completing one circuit, the distance has increased.}\
\
textbf{(c)} quad &text{When completing one circuit, the displacement has decreased.}\
\
textbf{(d)} quad &boxed{d=28 text{m}}\
&boxed{Delta x=0}\
end{align*}
$$
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{3330 text{m}}{450 text{s}}\
&=quadboxed{7.4 frac{text{m}}{text{s}}}\
end{align*}
$$
begin{align*}
textbf{(a)} quad &text{Click to see the graph.}\
\
textbf{(b)} quad &boxed{v_{avg}=7.4 frac{text{m}}{text{s}}}\
end{align*}
$$
The equations of motion for object 1 and 2 are, respectively:
$$
begin{align*}
x_{1}&=5.4 text{m}+left(1.3 frac{text{m}}{text{s}}right)times t\
x_{2}&=8.1 text{m}-left(2.2 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
$$
begin{align*}
x_{1}=x_{2}\
end{align*}
$$
Rearranging the terms, we get the time when the objects collide:
$$
begin{align*}
5.4 text{m}+left(1.3 frac{text{m}}{text{s}}right)times t&=8.1 text{m}-left(2.2 frac{text{m}}{text{s}}right)times t\
left(1.3 frac{text{m}}{text{s}}+2.2 frac{text{m}}{text{s}}right)times t&=8.1 text{m}-5.4 text{m}\
rightarrowquad t&=dfrac{2.7 text{m}}{3.5 frac{text{m}}{text{s}}}\
&=0.77143 text{s}\
&=quadboxed{0.77 text{s}}\
end{align*}
$$
We used $textbf{the rule for multiplication and division for significant figures}$ when writing the final result.
$$
begin{align*}
x_{c}&=5.4 text{m}+1.3 frac{text{m}}{text{s}}cdot 0.77 text{s}\
&=5.4 text{m}+1.001 text{m}\
&=6.401 text{m}\
&=quadboxed{6.4 text{m}}\
end{align*}
$$
In this case, we used $textbf{the rule for addition and subtraction for significant figures}$ when writing the final result.
begin{align*}
textbf{(a)} quad &boxed{t=0.77 text{s}}\
\
textbf{(b)} quad &boxed{x_{c}=6.4 text{m}}\
end{align*}
$$
$$
begin{align*}
x_{1}=25 text{m}-left(5.6 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
However, we can’t do the same for the second object because we don’t know the magnitude of its velocity. The best we can do for now is:
$$
begin{align*}
x_{2}=13 text{m}+v_{2}t\
end{align*}
$$
$$
begin{align*}
x_{1}=x_{2}\
end{align*}
$$
Rearranging the terms and solving the above equation for $v_{2}$ we obtain:
$$
begin{align*}
25 text{m}-5.6 frac{text{m}}{text{s}}cdot 0.61 text{s}&=13 text{m}+0.61 text{s}times v_{2}\
0.61 text{s}times v_{2}&=12 text{m}-5.6cdot 0.61 text{m}\
rightarrowquad v_{2}&=dfrac{8.584 text{m}}{0.61 text{s}}\
&=14.072131 frac{text{m}}{text{s}}\
&=quadboxed{14 frac{text{m}}{text{s}}}\
end{align*}
$$
The equation of motion for the second object is
$$
begin{align*}
x_{2}&=13 text{m}+left(14 frac{text{m}}{text{s}}right)times t\
end{align*}
$$
Notice that we applied $textbf{the rule for multiplication and division for significant figures}$ when writing the final result for $v_{2}$.
Using the equation of motion of the first object, we get the point of collision:
$$
begin{align*}
x_{c}&=25 text{m}-5.6 frac{text{m}}{text{s}}cdot 0.61 text{s}\
&=25 text{m}-3.416 text{m}\
&=21.584 text{m}\
&=quadboxed{22 text{m}}\
end{align*}
$$
In this case, we applied $textbf{the rule for addition and subtraction for significant figures}$ when writing the final result.
begin{align*}
textbf{(a)} quad &boxed{v_{2}=14 frac{text{m}}{text{s}}}\
\
textbf{(b)} quad &boxed{x_{c}=22 text{m}}\
end{align*}
$$
Assume positions of the two objects be equal to each other at the time of collision, and use the given data to decide the velocity of object 2. Then use an equation of motion to hit upon the position of the two objects at the instant of collision. While it is not necessary for solving this problem, position-time graphs for the two objects are shown at the right to help you picture the problem.
$textbf{Solution :}$
Assume the final positions of the two objects equal to each other:
$$
x_{f,1}= x_{f,2}
$$
$$
x_{i,1}+v_1t= x_{i,2}+v_2t
$$
$$
x_{i,1}+v_1 – x_{i,2}=v_2t
$$
$$
v_2=frac{x_{i,1}+v_1 – x_{i,2}}{t}=frac{(25m-13m)-(2.6m/s)(0.61s)}{0.61s}=color{#4257b2} boxed{bf 14m/s}
$$
With the equation of motion we can find the final position of the object
$textbf{Solution :}$
$$
x_f=(25m)+(-5.6m/s)(0.61s)=color{#4257b2} boxed{bf 22m}
$$
v_2=14m/s;and;x_f=22m
$$
$$
begin{align*}
v_{avg,1}&=dfrac{x}{t}\
rightarrowquad x&=v_{avg,1}t\
&=12.0 frac{text{km}}{text{h}}times 15 text{min}cdotdfrac{1 text{h}}{60 text{min}}\
&=12.0 frac{text{km}}{text{h}}times 0.25 text{h}\
&=quadboxed{3.00 text{km}}\
end{align*}
$$
The remaining distance to the church is
$$
begin{align*}
Delta x&=17.0 text{km}-3.00 text{km}\
&=quadboxed{14.0 text{km}}\
end{align*}
$$
and the remaining time before the ceremony is $Delta t=15 text{min}=0.25 text{h}$.
To make it on time, the average velocity should be
$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}\
&=dfrac{14.0 text{km}}{0.25 text{h}}\
&=quadboxed{56.0 frac{text{km}}{text{h}}}\
end{align*}
$$
When writing the results, we used $textbf{the rules for mathematical operations for significant figures.}$
begin{align*}
boxed{v_{avg}=56.0 frac{text{km}}{text{h}}}\
end{align*}
$$
Answers will depend and change as per situation. The San Andreas fault slips at an average speed of 35 mm per year. The cities of Los Angeles and San Francisco are separated by 560 km. The time for the fault to slip by 560 km is
$textbf{Solution :}$
$$
t=frac{Delta x}{v_{avg}}=frac{560km}{35mm/yr}frac{1000mm}{m}frac{1000m}{km}=color{#4257b2} boxed{bf 1.6times 10^7 hr}
$$
Or about 16 million years
1.6times 10^7 hr
$$
$$
x = left( 1.2~mathrm{m/s} right)t
$$
$$
x = 2.0~mathrm{m} + left( 0.70~mathrm{m/s} right)t
$$
$$
begin{align*}
v&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{2.0 text{m}}{2.0 text{s}}\
&=quadboxed{1.0 frac{text{m}}{text{s}}}\
end{align*}
$$
Therefore, the correct answer is $textbf{B.}$
begin{align*}
textbf{B.} quad 1.0 frac{text{m}}{text{s}}\
end{align*}
$$
The slope of the line corresponds to the velocity. Determine the slope of the line between $t=0s$ to $t=2s$, the time interval during which the robot is walking forward. The slope of the line is the velocity of the robot during that time interval.
$textbf{Solution :}$
$$
v_{av}=frac{rise}{run}=frac{Delta x}{Delta t}=frac{2-0m}{2-0s}=color{#4257b2} boxed{bf 1.0m/s}
$$
v_{av}=1.0m/s
$$
$$
begin{align*}
v&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{1.0 text{m}-2.0 text{m}}{8.0 text{s}-5.0 text{s}}\
&=dfrac{-1.0 text{m}}{3.0 text{s}}\
&=quadboxed{-0.33 frac{text{m}}{text{s}}}\
end{align*}
$$
Therefore, the correct answer is $textbf{A.}$
begin{align*}
textbf{A.} quad -0.33 frac{text{m}}{text{s}}\
end{align*}
$$
Calculate the slope of the line between $t=5s$ to $t=8s$, the time interval during which the robot is walking backward. The slope of the line is the velocity of the robot during that time interval.
$textbf{Solution :}$
$$
v_{av}=frac{rise}{run}=frac{Delta x}{Delta t}=frac{1-2m}{8-5s}=color{#4257b2} boxed{bf -0.33m/s}
$$
v_{av}=-0.33m/s
$$
Find the change in position of the robot between the times t = 2 s and t = 8 s. The displacement is the change in position over that time interval.
$textbf{Solution :}$
Find the change in position. The calculated result is answer choice B
$$
Delta x = 1m-2m=color{#4257b2} boxed{bf -1.0m}
$$
Delta x = -1.0m
$$
$$
begin{align*}
Delta x&=x_{f}-x_{i}\
&=1.0 text{m}-2.0 text{m}\
&=quadboxed{-1.0 text{m}}\
end{align*}
$$
which means that the correct answer is $textbf{B.}$
begin{align*}
textbf{B.} quad -1.0 text{m}\
end{align*}
$$
$$
begin{align*}
v&=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
&=dfrac{1 text{m}-2 text{m}}{8.0 text{s}-2.0 text{s}}\
&=dfrac{-1 text{m}}{6.0 text{s}}\
&=quadboxed{-0.17 frac{text{m}}{text{s}}}\
end{align*}
$$
Therefore, the correct answer is $textbf{A.}$
begin{align*}
textbf{A.} quad -0.17 frac{text{m}}{text{s}}\
end{align*}
$$
Refer to the given graph and determine the slope. Use the displacement together with the elapsed time to find the average velocity of the robot.
$textbf{Solution :}$
Find the average velocity. The calculated result is answer choice A.
$$
v_{av}=frac{Delta x}{Delta t}=frac{-1.0m}{8-2s}=color{#4257b2} boxed{bf -0.17m/s}
$$
v_{av}=-0.17m/s
$$
