All Solutions
Page 39: Practice Problems
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.
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.
Both Pedestrians were on Broad Street the entire time.
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.
(29 position intervals) $times$ (2.0 seconds/interval) = $sim$58 seconds.
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.
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.
We find the $20text{ s}$ mark and draw a vertical line at that point.
This line will intersect both of the runner’s trajectories.
Thus we conclude that at the given moment of time, the distance between the two runners is $d=25text{ m}$.
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!
– 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.
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:
$$
t = 0.1text{ h} cdot frac{60text{ min}}{h} = 6text{ min}
$$
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.
b) No(no intersection)
