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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 91: Lesson Check

Exercise 34

Step 1

1 of 2

In this problem, we find the relationship between the average velocity and the initial and final velocities of an object traveling with constant acceleration.

Step 2

2 of 2

If the acceleration is constant, we can write
$$
v_text{av} = frac{v_text{f} + v_text{i}}{2}
$$
We see that the average speed is the $textbf{arithmetic mean}$ of the initial and final velocities of an object traveling with constant acceleration.

Exercise 35

Step 1

1 of 2

From the calculus point of view, the area under the velocity-time curve is equal to the distance the object has travelled. Thus, the area under this object’s velocity-time curve is equal to the 5 m distance that it travels.

Result

2 of 2

Click here to see the explanation.

Exercise 36

Step 1

1 of 2

The rate of change of velocity is called acceleration. Hence, if the acceleration of a train is increased, the rate at which its velocity changes will also increase.

Result

2 of 2

Click here to see the explanation.

Exercise 37

Step 1

1 of 3

In this problem, applying breaks to a car gives it a constant acceleration. If the driving speed is doubled, we find what happen to the stopping time and stopping distance of the car.

Step 2

2 of 3

Part A.

We relate the stopping time $t$ to the initial driving velocity $v_text{i}$. The final velocity must be $v_text{f} = 0$ since the car stops.

$$
begin{align*}
v_text{f} &= v_text{i} + at \
0 &= v_text{i} + at \
implies t &= left( -frac{1}{a} right)v_text{i} \
t &propto v_text{i}
end{align*}
$$

The time it takes for the car to the driving speed when the driver hits the breaks. Hence, if the driving speed is doubled, the stopping time must also $textbf{doubled}$.

Step 3

3 of 3

Part B.

We now relate the stopping distance $Delta x$ to the initial velocity $v_text{i}$

$$
begin{align*}
v_text{f}^{2} &= v_text{i}^{2} + 2a Delta x \
0 &= v_text{i}^{2} + 2a Delta x \
implies Delta x &= left( -frac{1}{2a} right)v_text{i}^{2} \
Delta x &propto v_text{i}^{2}
end{align*}
$$

The distance needed to stop is proportional to the square of the driving velocity when the driver hits the breaks. The stopping distance must have $textbf{quadrupled}$.

Exercise 38

Step 1

1 of 3

In this problem, the observer sees the red light of a traffic light and hits the breaks. The acceleration is constant until the vehicle stops. The initial speed is $v_text{i} = 12~mathrm{m/s}$. We find the average speed.

Step 2

2 of 3

First, we find the average velocity.

$$
begin{align*}
v_text{av} &= frac{v_text{i} + v_text{f}}{2} \
&= frac{12~mathrm{m/s} + 0}{2} \
v_text{av} &= 6.0~mathrm{m/s}
end{align*}
$$

The average speed is the magnitude of the average velocity, so we have
$$
boxed{ s_text{av} = 6.0~mathrm{m/s} }
$$

Result

3 of 3

$$
s_text{av} = 6.0~mathrm{m/s}
$$

Exercise 39

Step 1

1 of 4

In this problem, a baseball player is dashing towards a home plate with a speed of $v_text{i} = 5.8~mathrm{m/s}$. She slides for $t = 1.1~mathrm{s}$ until stop ($v_text{f} = 0$). We find her acceleration and stopping distance.

Step 2

2 of 4

Part A.

To calculate the acceleration, we have

$$
begin{align*}
v_text{f} &= v_text{i} + at \
implies a &= frac{v_text{f} – v_text{i}}{t} \
&= frac{0 – 5.8~mathrm{m/s}}{1.1~mathrm{s}} \
&= -5.27273~mathrm{m/s^{2}} \
a &= boxed{ -5.3~mathrm{m/s^{2}} }
end{align*}
$$

Step 3

3 of 4

Part B.

For the distance she covers to stop, we have

$$
begin{align*}
v_text{f}^{2} &= v_text{i}^{2} + 2a Delta x \
implies Delta x &= frac{v_text{f}^{2} – v_text{i}^{2}}{2a} \
&= frac{v_text{f}^{2} – v_text{i}^{2}}{2 left( frac{v_text{f} – v_text{i}}{t} right)} \
&= frac{v_text{f} + v_text{i}}{2} t \
&= frac{0 + 5.8~mathrm{m/s}}{2} left( 1.1~mathrm{s} right) \
&= 3.19~mathrm{m} \
Delta x &= boxed{ 3.2~mathrm{m} }
end{align*}
$$

Result

4 of 4

begin{enumerate}
item [a)] $a = -5.3~mathrm{m/s^{2}}$
item [b)] $Delta x = 3.2~mathrm{m}$
end{enumerate}

Exercise 40

Step 1

1 of 2

$$
textbf{Concept :}
$$

With the help of velocity-position equation for constant acceleration we can determine the velocity of the firefighter after sliding a known distance.

Compute the final velocity from velocity-position equation

$$
textbf{Solution :}
$$

$$
v^2_f= v^2_i+2aDelta x
$$

$$
v_f=sqrt{ v^2_i+2aDelta x}=sqrt{0^2+2(6.2m/s^2)(3.7m)}=sqrt{45.88m/s^2}=color{#4257b2} boxed{bf 6.8m/s}
$$

Result

2 of 2

$$
v_f=6.8m/s
$$

Exercise 41

Step 1

1 of 6

In this problem, we are given the velocity-time graph of an ice skater. We find the distance covered by the skater between a) $t = 0$ to $t = 2~mathrm{s}$ and b) $t = 2~mathrm{s}$ to $t = 5~mathrm{s}$.

Step 2

2 of 6

Part A.

The distance covered is the product of the average speed and time interval for constant acceleration region. In this time interval, the slope is constant o the acceleration is also constant.

$$
begin{align*}
Delta x_{0 to 2} &= left( frac{v_text{i} + v_text{f}}{2} right) Delta t \
&= left( frac{0 + 2~mathrm{m/s}}{2} right) left( 2~mathrm{s} right) \
Delta x_{0 to 2} &= boxed{ 2.00~mathrm{m} }
end{align*}
$$

Step 3

3 of 6

Part B.

We split this region into the two intervals of constant acceleration. For time interval $t = 2mathrm{s}$ to $t = 4~mathrm{s}$

$$
begin{align*}
Delta x_{2 to 4} &= left( frac{v_text{i} + v_text{f}}{2} right) Delta t \
&= left( frac{2~mathrm{m/s} + 2.5~mathrm{m/s}}{2}right) left( 2~mathrm{s} right) \
Delta x_{2 to 4} &= 4.50~mathrm{m}
end{align*}
$$

Step 4

4 of 6

For time interval $t = 4mathrm{s}$ to $t = 45~mathrm{s}$

$$
begin{align*}
Delta x_{2 to 4} &= left( frac{v_text{i} + v_text{f}}{2} right) Delta t \
&= left( frac{2.5~mathrm{m/s} + 1~mathrm{m/s}}{2}right) left( 1~mathrm{s} right) \
Delta x_{2 to 4} &= 1.75~mathrm{m}
end{align*}
$$

Step 5

5 of 6

The total distance for the time interval $t = 2mathrm{s}$ to $t = 5~mathrm{s}$ must be

$$
begin{align*}
Delta x_{2 to 5} &= Delta x_{2 to 4} + Delta x_{4 to 5} \
&= 4.50~mathrm{m} + 1.75~mathrm{m}\
Delta x_{2 to 5} &= boxed{ 6.25~mathrm{m} }
end{align*}
$$

Result

6 of 6

begin{enumerate}
item [a)] $Delta x_{0 to 2} = 2.00~mathrm{m}$
item [b)] $Delta x_{2 to 5} = 6.25~mathrm{m}$
end{enumerate}

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