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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 62: Lesson Check

Exercise 40

Step 1

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The position-time equation of motion at a constant velocity has the following form:

$$
begin{align*}
x_{f}=x_{i}+vt\
end{align*}
$$

where $x_{i}$ and $x_{f}$ are the initial and final position, respectively. This simple linear equation has more to it than just giving you the final position. In fact, $textbf{provided that the velocity and the initial position of an object are known, the position-time equation gives the object’s position $x$ at any time $t$.}$

Result

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On condition that the speed and the initial position are known, the position-time equation can be used to determine an object’s position $x$ at any time $t$ during motion.

Exercise 41

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The equation of motion with constant velocity is a linear equation, i.e., an equation for a straight line. Such equations are often written in the slope-intercept form,

$$
y=ax+b
$$

where $a$ is the slope, and $b$ is the $y$ intercept.

Looking at the equation of motion

$$
x_{f}=x_{i}+vt
$$

it’s obvious that it’s actually the same as the one above, but with rearranged terms and different names of variables. The slope of the line $v$ represents velocity, and the $y$ intercept represents the initial position $x_{i}$.

Result

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Graphically, the equation of motion is represented by a straight line. The slope of the line represents the constant velocity $v$ at which an object moves, and the $y$ intercept of the line represents its initial position $x_{i}$.

Exercise 42

Solution 1

Solution 2

Step 1

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Both equations are written in the form of $x=x_{i}+vt$, so we can easily see that the velocity of each bike is

$$
v_{1}=2.5 frac{text{m}}{text{s}}
$$

$$
v_{2}=-3.5 frac{text{m}}{text{s}}
$$

$textbf{Knowing that speed is the magnitude (or absolute value) of velocity, we conclude that the second bike has the greater speed.}$

Result

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The second bicycle has the greater speed.

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We have been provided with the equations of motion for two bicycles and that for each of this equation, velocity is being multiplied by time concluding that bicycle 1 has a velocity of 2.5 m/s and bicycle 2 has a velocity of -3.5m/s. The speeds of each bicycle are the amount of these velocities. We conclude that $text{color{#4257b2} boxed{bf bicycle 2} has the greater speed}$

Result

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Click here to see the explanation.

Exercise 43

Step 1

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Plugging $t=0$ into both equations, we get the positions of bicycles:

$$
begin{align*}
x_{1}&=-4 text{m}\
x_{2}&=6.7 text{m}\
end{align*}
$$

As discussed before, distance has no association whatsoever with direction. Therefore, the distance between the bicycles at $t=0$ is

$$
begin{align*}
d&=4 text{m}+6.7 text{m}\
&=quadboxed{10.7 text{m}}\
end{align*}
$$

If this is confusing, think of it this way: the first bicycle is at $x_{1}=-4 text{m}$, and the second is at $x_{2}=6.7 text{m}$. The distance from the first bike to the origin (where $x=0$) is $4 text{m}$, and the distance from the origin to the second bicycle is $6.7 text{m}$. Thus, the distance between the two bikes is those two distances added, which $10.7 text{m}$.

Step 2

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At $t=1 text{s}$, the positions of bicycles are

$$
begin{align*}
x_{1}&=-1.5 text{m}\
x_{2}&=3.2 text{m}\
end{align*}
$$

The distance between them decreases because they travel in opposite directions, or, towards one another. In this case, the distance between them is

$$
begin{align*}
d&=1.5 text{m}+3.2 text{m}\
&=quadboxed{4.7 text{m}}\
end{align*}
$$

Result

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$circ$ The distance at $t=0$: $boxed{d=10.7 text{m}}$

$circ$ The distance at $t=1 text{s}$: $boxed{4.7 text{m}}$

Exercise 44

Solution 1

Solution 2

Step 1

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Plugging in the value of $t$, we quickly obtain the result:

$$
begin{align*}
x&=3.0 text{m}-5.0 frac{text{m}}{text{s}}times 1.5 text{s}\
&=3.0 text{m}-7.5 text{m}\
&=quadboxed{-4.5 text{m}}\
end{align*}
$$

Result

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$$
begin{align*}
boxed{x=-4.5 text{m}}\
end{align*}
$$

Step 1

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$$
textbf{Concept:}
$$

The equation in the linear form look like $x_t=x_i+vt$ where time $t$ is multiplied by velocity $v$, and the answer is added to the initial position. We will use the given equation to find the velocity together with initial speed of the bowling ball.

$$
textbf{Solution:}
$$

Wit the given equation and inserting numerical values we can find the final position of the ball:

$$
x_f=(3.0m)+(-5.0m/s)t
$$

$$
=(3.0m)+(-5.0m/s)(1.5)=color{#4257b2} boxed{bf -4.5m}
$$

Result

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$$
-4.5m
$$

Exercise 45

Solution 1

Solution 2

Step 1

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$textbf{(a)}$ Since both equations are written in the form of

$$
begin{align*}
x_{f}=x_{i}+vt
end{align*}
$$

we can easily see that the velocity of each bumper car is

$$
begin{align*}
v_{1}&=1.5 dfrac{text{m}}{text{s}}\
v_{2}&=-2.5 dfrac{text{m}}{text{s}}\
end{align*}
$$

Despite having a negative velocity, $textbf{the second car has the greater speed, because speed is the magnitude (or absolute value) of velocity,}$

$$
begin{align*}
boxed{abs{v_{2}}>abs{v_{1}}}
end{align*}
$$

Step 2

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$textbf{(b)}$ In case of collision, $textbf{the bumper cars have the same final position:}$

$$
begin{align*}
x_{1}&=x_{2}\
x_{i,1}+v_{1}t&=x_{i,2}+v_{2}t\
end{align*}
$$

Plugging in the values and solving for $t$ gives:

$$
begin{align*}
-4.0 text{m}+left(1.5 dfrac{text{m}}{text{s}}right)t&=8.8 text{m}-left(2.5 dfrac{text{m}}{text{s}}right)t\
left(1.5 dfrac{text{m}}{text{s}}+2.5 dfrac{text{m}}{text{s}}right)t&=8.8 text{m}+4.0 text{m}\
rightarrowquad t&=dfrac{12.8 text{m}}{4.0 frac{text{m}}{text{s}}}\
&=quadboxed{3.2 text{s}}\
end{align*}
$$

Result

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$$
begin{align*}
textbf{(a)}quad &boxed{abs{v_{2}}>abs{v_{1}}}\
\
\
\
textbf{(b)}quad &boxed{t=3.2 text{s}}\
end{align*}
$$

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$tt {(a) textbf{Speed} is the magnitude of velocity: $ Speed =||vec{v}||ge 0$}$

given the equations of the two bumper cars, the second car has a greater speed than the first one:
$$
2.5m>1.5m
$$

Step 2

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$$
tt{(b)Supposing that the two cars collide at point $A (x_A,t_A)$, we can write the two cars equations as follows at the time of collision:}
$$

$$
begin{align}
x_A=(-4.0m)+(1.5frac{m}{s})t_A\
x_A=(8.8m)+(-2.5frac{m}{s})t_A
end{align}
$$

$$
tt{we equalize now the two second members of equation(1) and (2) to find the time of collision:}
$$

$$
begin{align*}
(-4.0m)+(1.5frac{m}{s})t_A &=x_A=(8.8m)+(-2.5frac{m}{s})t_A\
(4frac{m}{s})t_A&=12.8(m)\
t_A&=frac{(12.8m)}{(4frac{m}{s})}\
&=boxed{textcolor{#4257b2}{3.2s}}
end{align*}
$$

Result

3 of 3

$$
tt{(a) the second car has greater speed, (b) 3.2s}
$$

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