Home Page Textbooks Physics Page 52: Practice Problems

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 52: Practice Problems

Exercise 14

Step 1

1 of 2

The average velocity is defined as displacement per unit time, or

$$
begin{align*}
v_{avg}=dfrac{Delta x}{Delta t}=dfrac{x_{f}-x_{i}}{t_{f}-t_{i}}\
end{align*}
$$

In this case, the average velocity is negative, which means that our athlete moves in the negative direction, towards the starting line. Using the equation above, solving for $Delta t$, and plugging in the values, we get

$$
begin{align*}
Delta t&=dfrac{Delta x}{v_{avg}}\
&=dfrac{x_{f}-x_{i}}{v_{avg}}\
&=dfrac{0 text{m}-50.0 text{m}}{-1.50 frac{text{m}}{text{s}}}\
&=quadboxed{33.3 text{s}}\
end{align*}
$$

Result

2 of 2

$$
begin{align*}
boxed{Delta t=33.3 text{s}}\
end{align*}
$$

Exercise 15

Solution 1

Solution 2

Step 1

1 of 2

$textbf{Velocity is a vector, meaning that it has both magnitude and direction.}$ Two velocities are the same only if they have the same magnitude and direction. Thus, if the two tennis players run towards the net having the same speed, $textbf{their velocities aren’t equal because they differ in direction.}$

Result

2 of 2

No, their velocities are not equal because they differ in direction.

Step 1

1 of 2

If two tennis players run with the same speed but in opposite directions, no, their velocities are not equal because velocity is a vector that includes both speed and direction. Two velocities can only be equal if their speeds and directions are the same

Result

2 of 2

Click here to see the explanation.

Exercise 16

Solution 1

Solution 2

Step 1

1 of 3

$textbf{(a)}$ The person’s average speed during the adventure is the total distance travelled (which is the circumference of the Earth) divided by elapsed time:

$$
begin{align*}
text{average speed}&=dfrac{text{distance}}{text{elapsed time}}\
&=dfrac{40 075 text{km}}{80 text{days}cdotleft(frac{24 text{h}}{1 text{day}}right)}\
&=dfrac{40 075 text{km}}{1920 text{h}}\
&=quadboxed{20.9 frac{text{km}}{text{h}}}\
end{align*}
$$

Step 2

2 of 3

$textbf{(b)}$ $textbf{The average velocity for the entire trip is zero,}$ the reason being that the final and initial position are the same.

$$
begin{align*}
boxed{v_{avg}=0 frac{text{km}}{text{h}}}\
end{align*}
$$

Result

3 of 3

$$
begin{align*}
textbf{(a)} quad &boxed{text{average speed}=20.9 frac{text{km}}{text{h}}}\
\
\
\
textbf{(b)} quad &boxed{v_{avg}=0 frac{text{km}}{text{h}}}\
end{align*}
$$

Step 1

1 of 3

a.) $textbf{Concept:}$
Use the definitions of average speed and average velocity to answer the questions.

$$
textbf{Solution:}
$$

Find the average speed, noting the distance travelled is the 40,075-km circumference of the Earth:

$$
Avg.,Speed=frac{distance}{time,passed}=frac{40075km}{80cancel{d}}times{1cancel{d}}{24h}=color{#4257b2} boxed{bf 20.9km/hr}
$$

Step 2

2 of 3

b.) $textbf{Concept:}$
Because the initial and final positions are the same, hence $Delta x = 0$:

$$
textbf{Solution:}
$$

$$
v_{av}=frac{Delta x}{Delta t}=color{#4257b2} boxed{bf 0}
$$

Result

3 of 3

$$
Avg.,Speed=20.9km/hr;and;v_{av}=0
$$

Exercise 17

Solution 1

Solution 2

Step 1

1 of 4

To solve this problem, we will calculate the average velocity of each train individually.

Step 2

2 of 4

Train A travels north, which is the positive direction, and covers 10 meters in one second. The average velocity of train A is then

$$
begin{align*}
v_{avg,A}&=dfrac{10 text{m}}{1 text{s}}\
&=quadboxed{10 frac{text{m}}{text{s}}}\
end{align*}
$$

Train B heads south, which is the negative direction, and covers 900 meters in 60 seconds. Therefore, the average velocity of train B is

$$
begin{align*}
v_{avg,B}&=dfrac{-900 text{m}}{60 text{s}}\
&=quadboxed{-15 frac{text{m}}{text{s}}}\
end{align*}
$$

Train C moves south with a speed of $20 frac{text{m}}{text{s}}$. It’s average velocity is

$$
begin{align*}
v_{avg,C}&=dfrac{-20 text{m}}{1 text{s}}\
&=quadboxed{-20 frac{text{m}}{text{s}}}\
end{align*}
$$

Finally, train D moves in the positive direction, and it’s average velocity is

$$
begin{align*}
v_{avg,D}&=dfrac{24 text{m}}{2 text{s}}\
&=quadboxed{12 frac{text{m}}{text{s}}}\
end{align*}
$$

Step 3

3 of 4

Ranking the trains from most negative to most positive velocity, we have

$$
begin{align*}
boxed{v_{avg,C}<v_{avg,B}<v_{avg,A}<v_{avg,D}}\
end{align*}
$$

Result

4 of 4

$$
begin{align*}
boxed{v_{avg,C}<v_{avg,B}<v_{avg,A}<v_{avg,D}}\
end{align*}
$$

Step 1

1 of 2

$$
textbf{Concept:}
$$

We need to calculate the average velocity of the train for each case, assuming north be the positive direction and then compare the velocities to rank them from the most negative to the most positive.

$$
textbf{Solution:}
$$

Calculate the average velocity of train A:

$$
v_{av,A}=frac{Delta x_A}{Delta t}=frac{10m}{1.0s}=color{#4257b2} boxed{bf 10m/s}
$$

Repeat the process for B, C and D we get:

$$
v_{av,B}=frac{Delta x_B}{Delta t}=frac{-900m}{60 s}=color{#4257b2} boxed{bf -15m/s}
$$

$$
v_{av,C}=frac{Delta x_C}{Delta t}=frac{-2times 10m}{1.0s}=color{#4257b2} boxed{bf -20m/s}
$$

$$
v_{av,D}=frac{Delta x_D}{Delta t}=frac{+24m}{2.0s}=color{#4257b2} boxed{bf +12m/s}
$$

Now by comparison we conclude that…

$$
color{#4257b2} boxed{bf v_{av,c}<v_{av,B}<v_{av,A}<v_{av,D}}
$$

Result

2 of 2

$$
v_{av,c}<v_{av,B}<v_{av,A}<v_{av,D}
$$

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Page 52: Practice Problems

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