All Solutions
Page 85: Standardized Test Practice
textbf{underline{textit{Solution}}}
$$
[ d= v_i t + dfrac{1}{2}at^2tag{1}]
Where,
newenvironment{conditions}
{parvspace{abovedisplayskip}
noindent
begin{tabular}{>{$}c{${}}c<{{}$} @{} p{9.1 cm}}}
{end{tabular}parvspace{belowdisplayskip}}
begin{conditions}
d &: & The distance traveled by the ball.\
v_i & :& The initial velocity of the ball .\
a &: & The acceleration constant of the ball.\
t &: & The duration of the motion of the ball.\
end{conditions}
And it is given that the constant of acceleration is 2 m/s$^2$, and that the duration of motion is 4 seconds.\
$$
begin{align*}
d &= 0times 4 + dfrac{1}{2} times 2.0 times 4^2\
&= 0.5 times 2.0 times (4.0)^2\
&=fbox{$16 ~ rm{m}$}
end{align*}
$$
$textbf{underline{textit{note:}}}$ Since the least number of significant figures in the multiplication operation is 2, therefor the final result must have at least 2 significant figures thus the answer is 16 m.
a = dfrac{v_{f}-v_{i}}{t}
$$
2.0m/s^2 = dfrac{v_{f}-0}{4.0s}
$$
v_{f} = 8m/s
$$
v_{f}^{2} = v_{i}^{2} + 2ad
$$
(110km/h)^{2} = (80km/h)^{2} + 2a(.5km)
$$
a = 5700km/h^2 = .44m/s^2
$$
d=v_{i}t + dfrac{1}{2}at^{2}
$$
(85)=(0)t + dfrac{1}{2}(9.8)t^{2}
$$
distance traveled by the falling pot = 85m
initial velocity = $0 ms^{-2}$ since the pot was initially at rest.
acceleration = $9.8 ms^{-2}$ since the pot is falling freely hence acceleration is simply the gravity acting on the object.
time is the unknown to be calculated.
t = sqrt{dfrac{(2times85)}{9.8}} = 4.2s
$$
– Displacement: $d=85mathrm{~m}$
**Objective**
– Find the time of travel $t$.
In order to solve this problem, we need to relate the time of travel $t$ to the known information (Displacement $d$ and time of travel $t$). To do this, we can use the kinematic formula:
$$begin{align}
d=v_0cdot t+dfrac{acdot t^2}{2}
end{align}$$
Where $v_0$ is the initial velocity.
$$begin{align}
d&=v_0cdot t+dfrac{acdot t^2}{2} nonumber
\&=0cdot t+dfrac{acdot t^2}{2} nonumber
\&=dfrac{acdot t^2}{2}
end{align}$$
Now we can use this equation, to get time of travel $t$.
$$begin{align}
t=sqrt{dfrac{2cdot d}{a}}
end{align}$$
Using this equation, and taking into account that acceleration due to gravity is $a=g=9.81mathrm{~dfrac{m}{s^2}}$ (downwards).
$$begin{aligned}
t&=sqrt{dfrac{2cdot d}{a}}
\&=sqrt{dfrac{2cdot 85}{9.81}}
\&=4.16approx4.2mathrm{~sec}
end{aligned}$$
So the answer is $boxed{mathrm{A}:t=mathrm{4.2}mathrm{~sec}}$
$y = dfrac{1}{2} (9.8) (3.20)^2 + (0) (3.20)$
$$
y = 50.0 m
$$
50.0 m
$$
– Time of travel: $t=3.20mathrm{~sec}$
**Obejctive**
– Find the height (displacement of rock) $d$.
In order to solve this problem, we need to relate the displacement $d$ to the known parameters (Time of travel $t$ and acceleration due to gravity $a=g=9.81mathrm{dfrac{m}{s^2}}$). To do this, we can use the kinematic formula:
$$begin{align}
d=v_0cdot t+dfrac{acdot t^2}{2}
end{align}$$
Where $v_0$ is the initial veloicty.
$$begin{align}
d&=v_0cdot t+dfrac{acdot t^2}{2} nonumber
\&=0cdot t+dfrac{acdot t^2}{2} nonumber
\&=dfrac{acdot t^2}{2}
end{align}$$
$$begin{align}
d&=dfrac{acdot t^2}{2} nonumber
\&=dfrac{9.81cdot 3.2^2}{2} nonumber
\&approx 50mathrm{~m}
end{align}$$
So, the answer is $boxed{mathrm{C}: d=50mathrm{~m}}$.
$$
d = dfrac{v^2}{2 a} = dfrac{(25.278)^2}{2*6.40} = 50.0 m
$$
– Acceleration (deceleration) due to breaks: $a=-6.4mathrm{~dfrac{m}{s^2}}$
– Initial velocity: $v_0=91mathrm{~dfrac{km}{h}}=91cdotdfrac{1000}{3600}mathrm{~dfrac{m}{s}}=25.27mathrm{dfrac{m}{s}}$
**Objective**
– Find the distance of travel $d$ due to deceleration $a$.
In order to solve this problem, we need to relate the distance of travel $d$ to the known parameters (Deceleration $a$ and initial velocity $v_0$). To do this, we can use the kinematic formula:
$$begin{align}
v_f^2-v_0^2=2cdot a cdot d
end{align}$$
Where $v_f$ is the final velocity.
$$begin{align}
v_f^2-v_0^2&=2cdot a cdot d nonumber
\0^2-v_0^2&=2cdot a cdot drightarrow d=-dfrac{v_0^2}{2cdot a}
end{align}$$
Now we can use this equation to find the displacement of the car $d$.
$$begin{align}
d&=-dfrac{v_0^2}{2cdot a} nonumber
\&=-dfrac{25.27^2}{2cdot (-6.4)} nonumber
\&=49.89mathrm{~m}approx50mathrm{~m}
end{align}$$
So the answer is $boxed{mathrm{C}: d=50mathrm{~m}}$.
$==> 2 a d = v_f^2 – v_i^2$
$$
==> a = dfrac{(v_f^2 – v_i^2)}{2 a}
$$
dfrac{(v_f^2 – v_i^2)}{2 a}
$$
We must first determine the velocities and time intervals then use kinematic formulas to calculate the distance traveled.
From the graph we can see that the change of direction is not instantaneous because there is some slope when changing speed. But we can count as if it were instant at time $t = 25 mathrm {s}$. Although this will give us a slightly larger shift to the north, the shift to the south will be greater and these two differences will be subtracted and this will not affect the final result.
Known:
$$
begin{align*}
v_1&=25 mathrm{m/s} \
v_2&=-25 mathrm{m/s} \
t_1&=25 mathrm{s} -0 mathrm{s}=25 mathrm{s} \
t_2&=45 mathrm{s}-25 mathrm{s}=20 mathrm{s} \
end{align*}
$$
Unknown:
$$
begin{align*}
x&=?
end{align*}
$$
$$
begin{align*}
s&=vcdot t tag{formula for the total displacement in motion const velocity} \
\
x&=v_1cdot t_1 + v_2cdot t_2 \
&=25 mathrm{m/s}cdot 25 mathrm{s} -25 mathrm{m/s}cdot 20 mathrm{s} \
&=boxed{125 mathrm{m}}
end{align*}
$$
Note that the final result is positive which means that the shift is in a northerly direction.
textbf{underline{textit{Solution}}}
$$
textbf{{textit{Definition}}}
$$
$textbf{underline{textit{Average acceleration:}}}$ It is the change in the velocity during a given interval.
$textbf{underline{textit{The instantaneous acceleration:}}}$ It is the rate of change of the velocity at a given instant of time, where the average acceleration and the instantane-
ous acceleration are the same given that the acceleration is constant through the given time interval.
ration, but this is not the case if the object is not uniformly accelerating.
Where in a velocity-time graph, the slope of the line representing the motion during a time interval in the graph yields the acceleration of the object under study.
Using any two points on the given straight line we actually are calculating the average acceleration between the two points “or the average acceleration during this specific time interval”.
Where the slope of the straight line using any two points having a coordinates $left( x_1,y_1right)$ and $left( x_2,y_2right)$, is given by the following equation
$$
m=dfrac{y_2 – y_1}{x_2 – x_1}
$$
And since in v-t graph the $y$-axis represents the velocity of the object, and the $x$-axis represents the time, therefor the slope of any given straight line would be the change in the velocity over the change in time “i.e duration of time during which the change occurred” which is by definition is the average acceleration.
However, if we want to find the instantaneous acceleration of the object during a given instant of time, would provide tricky if the acceleration is not constant “i.e. the line representing the motion is not a perfect straight line or more like a curvy line”.
Where we would need to find the tangent line passing through this given point “point on the graph representing the instant of time and velocity of the object at which we want to find the instantaneous acceleration” and then by finding the slope of this tangent line we are actually calculating the instantaneous acceleration of the object under study.
a=dfrac{triangle v}{triangle t}
$$
a=dfrac{65.7-8.10}{12.00-0.00}
$$
a=4.8m/s^{2}
$$
D=dfrac{1}{2}bh+lw
$$
D=dfrac{1}{2}(12)(65.7-8.10)+(12)(8.10)
$$
D=442.8m
$$
$$begin{align}
a=dfrac{v-v_0}{t}
end{align}$$
Where $v_0$ is the initial velocity, $v$ is the current velocity and $t$ is the time it took for velocity to change from $v_0$ to $v$.
$$begin{align}
a&=dfrac{v-v_0}{t} nonumber
\&=dfrac{65.7-8.1}{12} nonumber
\&=boxed{4.8mathrm{~dfrac{m}{s^2}}}
end{align}$$
$$begin{align}
v^2-v_0^2=2cdot acdot d
end{align}$$
From the first part of the problem, we know all the parameters above except $d$, so the only thing we need to do is express $d$ from equation $(3)$ and calculate.
$$begin{align}
d&=dfrac{v^2-v_0^2}{2cdot a} nonumber
\&=dfrac{65.7^2-8.1^2}{2cdot 4.8} nonumber
\&=boxed{442.8mathrm{~m}}
end{align}$$
$d=442.8mathrm{~m}$
