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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 81: Lesson Check

Exercise 11

Step 1

1 of 2

In this problem, we are asked if an object can accelerate with constant speed.

Step 2

2 of 2

The answer is $textbf{yes}$. If the acceleration is always perpendicular to the velocity, the magnitude of the velocity does not change, hence the speed remains constant.

Exercise 12

Step 1

1 of 2

In this problem, we describe the velocity-time graph of an object with constant acceleration.

Step 2

2 of 2

The slope of the velocity-time graph is the acceleration. When the acceleration is constant, the slope mut also be constant. The graph must be $textbf{linear}$.

Exercise 13

Step 1

1 of 3

In this problem, we are asked if a negative acceleration always means decreasing speed, and a positive acceleration always means an increase in speed.

Step 2

2 of 3

These two statements $textbf{does not}$ always happen. If the acceleration is negative, it is pointing in the negative direction. However, if the velocity is along the negative direction, the negative acceleration would make the velocity
more negative. Since the speed is the magnitude of the velocity, the speed must be increasing as the velocity becomes more negative.

Step 3

3 of 3

On the other hand, an object with positive acceleration and negative velocity experience a decrease in speed. The velocity becomes less negative, so its magnitude, the speed, must be decreasing.

Exercise 14

Step 1

1 of 2

In this problem, the change in velocity is constant, but the time interval decreases. We find what happens to the average acceleration.

Step 2

2 of 2

The average acceleration and time interval are related by

$$
begin{align*}
a_text{av} &= frac{Delta v}{Delta t} \
implies a_text{av} &propto frac{1}{Delta t}
end{align*}
$$

The relationship is inverse. If the time interval decreases, the acceleration must $textbf{increase}$.

Exercise 15

Step 1

1 of 3

In this problem, we are given the initial and final velocities, and time interval, of 4 situations. We compare their accelerations in increasing order
begin{center}
begin{tabular}{|c|c|c|c|c|}
hline
& A & B & C & D \ hline
$v_text{f}~(mathrm{m/s})$ & 5 & 25 & 25 & -15 \ hline
$v_text{i}~(mathrm{m/s})$ & 15 & 30 & 20 & -10 \ hline
$Delta t~(mathrm{s})$ & 10 & 15 & 2 & 3 \ hline
end{tabular}
end{center}

Step 2

2 of 3

The acceleration is the rate of change of the velocity. For each situation,

$$
begin{align*}
a_text{av} &= frac{Delta v}{Delta t} \
a_text{av, A} &= frac{5~mathrm{m/} – left( 15~mathrm{m/s} right)}{10~mathrm{s}} = -1.0~mathrm{m/s^{2}} \
a_text{av, B} &= frac{25~mathrm{m/} – left( 30~mathrm{m/s} right)}{15~mathrm{s}} = -0.33~mathrm{m/s^{2}} \
a_text{av, C} &= frac{25~mathrm{m/} – left( 20~mathrm{m/s} right)}{10~mathrm{s}} = 2.5~mathrm{m/s^{2}} \
a_text{av, D} &= frac{-15~mathrm{m/} – left( -10~mathrm{m/s} right)}{3~mathrm{s}} = -1.7~mathrm{m/s^{2}} \
end{align*}
$$

The order must be
$$
boxed{ a_text{av, D} < a_text{av, A} < a_text{av, B} < a_text{av, C} }
$$

Result

3 of 3

$$
a_text{av, D} < a_text{av, A} < a_text{av, B} < a_text{av, C}
$$

Exercise 16

Step 1

1 of 2

In this problem, we are asked if a ball with velocity $v = +1~mathrm{m/s}$ can possible have an acceleration of $a = -1~mathrm{m/s^{2}}$.

Step 2

2 of 2

The answer is $textbf{yes}$, since the acceleration is caused by an external force, and that force is not necessarily parallel to the velocity.

Exercise 17

Step 1

1 of 4

In this problem, a car initially moves with velocity $v_text{i} = 18~mathrm{m/s}$ due north. We let north be the positive direction. We find its velocity after a time interval $Delta t = 7.0~mathrm{s}$ if the acceleration is a) $a_text{av} = +1.5~mathrm{m/s^{2}}$ and b) $a_text{av} = -1.5~mathrm{m/s^{2}}$.

Step 2

2 of 4

Part A.

For this part, we have $a_text{av} = +1.5~mathrm{m/s^{2}}$

$$
begin{align*}
v_text{f} &= v_text{i} + a_text{av}Delta t \
&= 18~mathrm{m/s} + left( +1.5~mathrm{m/s^{2}} right) left( Delta t = 7.0~mathrm{s} right) \
&= 28.5~mathrm{m/s} \
v_text{f} &= boxed{ 28~mathrm{m/s} }
end{align*}
$$

Step 3

3 of 4

Part B.

For this part, we have $a_text{av} = -1.5~mathrm{m/s^{2}}$

$$
begin{align*}
v_text{f} &= v_text{i} + a_text{av}Delta t \
&= 18~mathrm{m/s} + left( -1.5~mathrm{m/s^{2}} right) left( Delta t = 7.0~mathrm{s} right) \
v_text{f} &= boxed{ 7.5~mathrm{m/s} }
end{align*}
$$

Result

4 of 4

begin{enumerate}
item [a)] $+28~mathrm{m/s}$
item [b)] $+7.5~mathrm{m/s}$
end{enumerate}

Exercise 18

Step 1

1 of 5

In this problem, we are given the velocity-time graph of a particle. We find the acceleration for the three time intervals.

Step 2

2 of 5

Part A.

For this part, $vtext{f} = 10~mathrm{m/s}$, $v_text{i} = 0~mathrm{m/s}$, $t_2 = 5~mathrm{s}$, and $t_{1} = 0~mathrm{s}$. The acceleration is

$$
begin{align*}
a_text{av} &= frac{Delta v}{Delta t} \
a_text{av, A}&= frac{v_text{f} – v_text{i}}{t_{2} – t_{2}} \
&= frac{10~mathrm{m/s} – left( 0~mathrm{m/s} right)}{5~mathrm{s} – 0~mathrm{s}} \
a_text{av, A} &= boxed{ 2~mathrm{m/s^{2}} }
end{align*}
$$

Step 3

3 of 5

Part B.

For this part, $vtext{f} = 10~mathrm{m/s}$, $v_text{i} = 10~mathrm{m/s}$, $t_2 = 15~mathrm{s}$, and $t_{1} = 5~mathrm{s}$. The acceleration is

$$
begin{align*}
a_text{av} &= frac{Delta v}{Delta t} \
a_text{av, B}&= frac{v_text{f} – v_text{i}}{t_{2} – t_{2}} \
&= frac{10~mathrm{m/s} – left( 10~mathrm{m/s} right)}{15~mathrm{s} – 5~mathrm{s}} \
a_text{av, B} &= boxed{ 0~mathrm{m/s^{2}} }
end{align*}
$$

Step 4

4 of 5

Part C.

For this part, $vtext{f} = 5~mathrm{m/s}$, $v_text{i} = 10~mathrm{m/s}$, $t_2 = 25~mathrm{s}$, and $t_{1} = 15~mathrm{s}$. The acceleration is

$$
begin{align*}
a_text{av} &= frac{Delta v}{Delta t} \
a_text{av, C}&= frac{v_text{f} – v_text{i}}{t_{2} – t_{2}} \
&= frac{5~mathrm{m/s} – left( 10~mathrm{m/s} right)}{25~mathrm{s} – 15~mathrm{s}} \
a_text{av, C} &= boxed{ -0.5~mathrm{m/s^{2}} }
end{align*}
$$

Result

5 of 5

begin{enumerate}
item [a)] $a_text{av, A} = 2~mathrm{m/s^{2}}$
item [b)] $a_text{av, B} = 0~mathrm{m/s^{2}}$
item [c)] $a_text{av, C} = -0.5~mathrm{m/s^{2}}$
end{enumerate}

Exercise 19

Step 1

1 of 3

In this problem, we are given that an object has average acceleration of $a_text{av} = +6.24~mathrm{m/s^{2}}$ after some time interval $Delta t = 0.300~mathrm{s}$. The final velocity is $v_text{f} = +9.31~mathrm{m/s}$. We find the initial velocity.

Step 2

2 of 3

We have the following equation

$$
begin{align*}
v_text{f} &= v_text{i} + a_text{av} Delta t \
implies v_text{i} &= v_text{f} – a_text{av} Delta t \
&= +9.31~mathrm{m/s} – left( +6.24~mathrm{m/s^{2}} right) left( 0.300~mathrm{s} right) \
&= 7.435~mathrm{m/s}\
v_text{i} &= boxed{ 7.44~mathrm{m/s} }
end{align*}
$$

Result

3 of 3

$$
v_text{i} = 7.44~mathrm{m/s}
$$

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Page 81: Lesson Check

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