Home Page Textbooks Physics: Principles and Problems Page 42: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

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Page 42: Section Review

Exercise 19

Step 1

1 of 4

$$
textbf{underline{textit{Solution}}}
$$

Step 2

2 of 4

The given particle model diagram, shows the motion of a baby crawling across the kitchen with time intervals of 1 seconds..\

And checking the given diagram carefully we notice that the points are equidistant which means that the baby is moving with a constant pace having a uniform speed.\

And the distance between 2 consecutive dots is 20 cm, thus the baby crawls a distance of 20 cm in 1 second, therefor the following textbf{table} describes the particle diagram
begin{center}
begin{tabular}{||c||c| c | c| c| c| c| c| c| c||}
hline hline
Position $left(rm{cm} right)$&0 & 20 & 40 & 60 & 80 & 100 & 120 &140 & 160\
hline
Time (s)& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\
hline hline
end{tabular}
end{center}

Step 3

3 of 4

From the previous table, where we plot the position on $y$-axis and the time on $x$-axis thus the $textbf{graph}$ as follows

Exercise scan

Result

4 of 4

See Graph.

Exercise 20

Solution 1

Solution 2

Step 1

1 of 3

$$
textbf{underline{textit{Solution}}}
$$

Step 2

2 of 3

From the given graph, we find that the graph is a relation between the position of the hockey puck as $y$-axis and the time of observation as $x$-axis, which means that the slope of this lines is the speed of the hockey puck and since the slope of the line is constant thus the speed of the hockey puck is constant, and the slope of the straight line can be calculated using the following equation

$$
m= dfrac{y_2-y_1}{x_2-x_1}tag{1}
$$

Where using any 2 points on the line and equation (1) we find that the slope of the straight line is 20 m/s, hence the hockey puck glides with a constant speed of 20 m/s.

Thus we can model such motion as equidistant dots “as the hockey pucks glides with a constant speed” and the distance between two successive points is 20 m such that the time interval between two consecutive dots is 1 seconds.

Therefor particle model $textbf{diagram}$ is as follows

Exercise scan

Result

3 of 3

Equidistant dots where the time interval between two consecutive dots is 1 seconds and the distance between them is 20 m, See Diagram.

Step 1

1 of 2

The time intervals between positions are 1 s. The positions show where the hockey puck is on a frozen pond over a span of 7 seconds.

Result

2 of 2

See Work for Graph.

Exercise 21

Step 1

1 of 4

$$
textbf{underline{textit{Solution}}}
$$

Step 2

2 of 4

From, the given graph, we draw a horizontal line from point (0,10) on $y$-axis and find the coordinated of point of intersection of this horizontal line with the straight line as in the following $textbf{graph}$.

Where we find that the $x$-coordinate of the point of intersection line is 0.5 s, thus the hockey puck was 10 m beyond the origin exactly at 0.5 sec.

Exercise scan

Step 3

3 of 4

$textbf{underline{textit{Another approach:}}}$ the slope of this line using any 2 points on the straight line and the following equation

$$
m=dfrac{y_2-y_1}{x_2-x_1}
$$

Is 20 m/s, and since the straight line is passing through the origin therefor the equation of this straight line is

$$
y=20 x tag{1}
$$

And since we want to find the exact moment at which the hockey puck is at a position 10 m beyond the origin, thus we substitute in the equation (1) by $y=10$ and calculate $x$, we get

$$
begin{align*}
10 &= 20 x\
x&= dfrac{10}{20} \
&= fbox{$0.5~ rm{s}$}
end{align*}
$$

Thus, the hockey goes 10 m beyond the origin at 0.5 seconds.

Result

4 of 4

At 0.5 seconds

Exercise 22

Step 1

1 of 4

$$
textbf{underline{textit{Solution}}}
$$

Step 2

2 of 4

From, the given graph, we draw a vertical line from points (0.5,0) on $x$-axis and find the coordinated of point of intersection of this vertical line with the straight line as in the following $textbf{graph}$.

Where we find that the $y$-coordinate of the point of intersection line is at a position 100 m beyond the origin, while at time 0 seconds the hockey puck was at the origin, thus in a time interval of 5 seconds the hockey puck cuts a distance of 100 m.

Exercise scan

Step 3

3 of 4

textbf{underline{textit{Another approach:}}} the slope of this line using any 2 points on the straight line and the following equation
[ m=dfrac{y_2-y_1}{x_2-x_1}]
Is 20 m/s, and since the straight line is passing through the origin therefor the equation of this straight line is

begin{align*}
y&=20 x tag{1}
end{align*}

And since we want to find the exact moment at which the hockey puck goes beyond 10 m, thus we substitute in the equation (1) by $x=0$ and $x=5$ and calculate $y$ at each time, we get
begin{align*}
intertext{At time $x=0$}
y_0 &= 20 x\
y_0&= 0 \
intertext{At time $x=5$}
y_5 &= 20 times 5\
&= 100 ~ rm{m}
intertext{Thus, the hockey in a time interval between 0 and 5 seconds cuts a distance }
d&=y_5 – y_0\
&= 100-0\
&= 100 ~ rm{m}
end{align*}

Thus, the hockey puck cuts a distance of 100 m in a time interval of 5 seconds.

Result

4 of 4

100 m

Exercise 23

Step 1

1 of 2

The puck was at position $d_i=40m$ at time $t_i=2.0s$

The puck was at position $d_f=80m$ at time $t_f=4.0s$

The time it took to change position from $d_i$ to $d_f$ is

$$
Delta t=t_f-t_i=2.0s
$$

Result

2 of 2

2.0 s

Exercise 24

Solution 1

Solution 2

Step 1

1 of 1

No, the two pictures aren’t describing the same motion. One picture shows an object moving quicker than the other even though both of them do show objects that are moving in a positive position.

Step 1

1 of 5

In this problem, we are asked to look at the given graph and the particle model and compare them.

We need to state whether they describe the same movement.

Step 2

2 of 5

**Looking at the particle model**

On the particle model, we see $8$ dots in total, so we have $7$ time intervals. We are told that each interval corresponds to $2text{ s}$.

We will look at the first $5$ time intervals(the first $6$ dots) as we can see that the total distance traveled during these $5$ intervals is $10text{ m}$.

Since we know that $10text{ m}$ is traveled during $5 cdot 2text{ s}=10text{ s}$, we conclude that:
$$
v_1 = frac{10text{ m}}{10text{ s}} = 1 , frac{text{m}}{text{s}}
$$

Step 3

3 of 5

**Looking at the graph**

On the graph, we see that a distance of $12text{ m}$ is traversed during a time interval of $3text{ s}$.

Using this we can find the velocity to equal:
$$
v_2 = frac{12text{ m}}{3text{ s}} = 4 , frac{text{m}}{text{s}}
$$

Step 4

4 of 5

**Conclusion**

In the previous two parts, we have found that even though both the particle model and the graph describe bodies moving uniformly, the velocity of the body on the particle model, $v_1$, is much smaller than the velocity corresponding to the body moving on the graph, $v_2$.

Result

5 of 5

No(different speeds)

unlock

All Solutions

Page 42: Section Review

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