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Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 39: Practice Problems

Exercise 9

Solution 1

Solution 2

Step 1

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$mathbf{Explanation:}$

In the distance time graph shown in figure 2-13, the car starts at 125.0 m from the origin and moves towards the origin, arriving at the origin after 5.0 seconds

The car keeps moving even after reaching the origin.

Result

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The car moves towards the origin.

Step 1

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In Figure 2-13, the car starts at 125.0m away from the origin. The car then moves towards and crosses the origin after five seconds. The car then moves beyond the origin.

Result

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Moves towards the origin

Exercise 10

Solution 1

Solution 2

Step 1

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Step 2

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The car displaces 125.0m in 5.0 seconds, and does so at a constant speed. The dots must be evenly spaced out and you must show an arrow pointing away TOWARDS the 0.0m mark (since the car is moving towards the origin).

Result

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See explanation for the diagram

Step 1

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In this problem, we are asked to look at the given graph and draw the corresponding motion diagram.

Step 2

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First, we will draw the origin and the $x$ axis with distances marked along with it.

As we can see from the graph, the car travels $25text{ m}$ every $1text{ s}$(since it uniformly travels $125text{ m}$ in $5text{ s}$)

We will mark each $25text{ m}$ with a dot, and do so for all $7text{ s}$ of movement(going from $+125text{ m}$ to $-50text{ m}$).

Finally, we will draw the vector arrow from the initial position to the final position.

Step 3

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The motion diagram is given bellow:

![‘slader’](https://slader-solution-uploads.s3.amazonaws.com/44863f00-98a3-4535-ac0f-4fae3a7a8bac-1649533358559902.png)

Exercise 11

Step 1

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A) Use Figure 2-13 to make an estimate for what time the car was 25.0m east of the origin. Go to about 25.0m up from the origin, then draw a line to the right. Your line will hit the graph’s green line at 4.0s. Therefore, the car was 25.0m east of the origin at about 4.0 seconds

Step 2

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B) Locate 1.0s along the bottom axis and draw a line straight up. Your line should cross the graph’s green line at around 100.0m. Therefore, the car is at about 100.0m east of the origin at 1.0 seconds.

Result

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A) 4.0s; B) 100.0m.

Exercise 12

Step 1

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Pedestrian A starts off west of High Street and walks to the east (which is the positive direction). Pedestrian A passes by Pedestrian B after some time, and then crosses the intersection between Broad Street and High Street a little bit after that.

Step 2

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Pedestrian B starts off east of High Street and walks to the west (which is the negative direction). Pedestrian B crosses High Street, and then passes Pedestrian A sometime after that.

Both Pedestrians were on Broad Street the entire time.

Result

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A passes B, then crosses High. B crosses High, then passes A.

Exercise 13

Step 1

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Odina walks 2.6m every 2.0 seconds. She starts from the Cafeteria to the Band Room, which is 100.0m away.

Think about this critically. After 10.0s (which is five time intervals), she will be five position intervals away from the Cafeteria (that is, 2.6 $times$ 5 = 13.0 m). This also means that after 20.0s, she will be 26.0m away from the Cafteria (ten time intervals and ten position intervals).

Without going into the formal definition of velocity, the amount of time it would take to reach 25.0m away from the cafeteria is a little bit less than 20 seconds; around 19 seconds, actually.

Step 2

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75.0m $div$ 2.6m = $sim$29 position intervals

(29 position intervals) $times$ (2.0 seconds/interval) = $sim$58 seconds.

Let’s approach the second part a little differently. 25.0m away from the Band Room would mean she’s 75.0m away from the Cafeteria; she walked 75.0m total.

We can figure out how many position intervals she has walked by dividing 75.0m by 2.6m. This equals about 29 position intervals.

If we multiply the number of position intervals by the 2.0 seconds/interval, we can get how much time it took to reach at 75.0m mark! It ends up being about 58 seconds.

Step 3

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The graph of Odina’s Motion will be a straight line. She starts at the origin, and she moves farther away as time goes on.

It’s a good idea to know what time she’ll reach the Band Room (that is, 100.0m away). Since 100.0m $div$ 2.6m = 38 position intervals, the amount of time to get to the Band Room is 38 $times$ 2.0s = 76 seconds.

So just graph a line using those two points: She starts at the origin, and reaches a position of 100.0m away after 76 seconds, as seen on the left.

Exercise 14

Step 1

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Example Problem 2 shows Runner A at the origin and Runner B at 50.0m away from the origin at t=0.0s. The big “event” that occurred was that Runner A crossed the origin at t=0.0s though!

Result

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Runner A crosses the origin.

Exercise 15

Step 1

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The graph in Example Problem 2 shows that Runner B passed Runner A after 45.0 seconds. Therefore, Runner B is ahead at 48.0 seconds.

Result

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Runner B.

Exercise 16

Step 1

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Looking at the graph of Example Problem 2: When Runner A was at 0.0m (i.e. the origin), the time was 0.0s. At that same time on the graph, Runner B was at -50.0 meters.

Result

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-50.0m

Exercise 17

Solution 1

Solution 2

Step 1

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To do this, you must draw a line upwards from the 20.0s mark on Example Problem 2’s graph. By doing so, you’ll find that Runner B was at about 50m away from the origin after 20.0s, and Runner A was about 80m away from the origin at 20.0s. Thus, the two runners are about 30m away from each other at t=20.0s.

Result

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About 30m apart.

Step 1

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In this problem, we are asked to look at the given graph and find the distance between the two runners at the moment of time $t=20.0text{ s}$.

Step 2

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Firstly we need to mark the moment of time on the graph.

We find the $20text{ s}$ mark and draw a vertical line at that point.

This line will intersect both of the runner’s trajectories.

Step 3

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Looking at the points where the vertical line intersects the lines of runners $A$ and $B$, we see that this distance is approximately half of the initial distance between the two runners marked on the $y$-axis$(50text{ m})$

Thus we conclude that at the given moment of time, the distance between the two runners is $d=25text{ m}$.

Result

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$d=25text{ m}$

Exercise 18

Solution 1

Solution 2

Step 1

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0.1 hours $times$ $dfrac{60min}{hour}$ = 6.0 minutes.

According to Figure 2-16, Heather starts walking 0.1 hours after Juanita started walking; Heather’s line starts at 0.1 hours. If you convert this to minutes, this ends up being 6.0 minutes.

Step 2

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Heather will never catch up to Juanita. The two are walking at constant speeds, and Heather is walking slower than Juanita. Since Juanita got a head start, Heather will just continue falling behind Juanita.

Graphically, this means that Juanita’s line and Heather’s line will never cross. That means Juanita and Heather will never meet at the same time and place!

Result

3 of 3

A) 6.0 minutes; B) No.

Step 1

1 of 4

In this problem we are asked to look at the given graph and find:
– The time interval Juanita had already spent walking when Heather began her walk.
– We also need to state whether Heather will ever catch up to Juanita.

Step 2

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**Part a)**

As we can see from the graph, Heather’s position starts changing at the moment of time corresponding to the first vertical line right of the $y$-axis.

We can see that $5$ intervals like this sum up to $0.5text{ h}$, thus this interval corresponds to:
$$frac{0.5text{ h}}{5}=0.1text{ h}$$

This is easily converted to minutes since one hour has $60$ minutes we write:
$$
t = 0.1text{ h} cdot frac{60text{ min}}{h} = 6text{ min}

Step 3

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**Part b)**

For Heather to catch up to Juanita, the two lines on the diagram need to intersect.

The point of intersection would represent the point in time and space where Heather caught up to her.

As we can see, the two lines do not intersect, and as time passes move further and further apart, so we conclude that Heather will never catch up to Juanita.

Result

4 of 4

a) $t=6text{min}$

b) No(no intersection)

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