Q#32
(II) A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/s when it passes a railway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Fig. 2–38.)
Answer:
Use the information for the first 180 m to find the acceleration and the information for the full motion to find the final velocity. For the first segment, the train has $v_0$ = 0 m/s, $v_1$ = 18 m/s, and a displacement of $x-x_0$ = −180 m. Find the acceleration from Eq.
$v_1^2=v_0^2+2a(x-x_0)$
$a=\frac{v^2-v_0^2}{2(x-x_0)}$
$a=\frac{(18 \ m/s)^2-(0)^2}{2(180 \ m)}=0.90 \ m/s^2$
Find the speed of the train after it has traveled the total distance (total displacement of $x-x_0$ = 225 m) using Eq.
$v_2^2=v_0^2+2a(x-x_0)$
$v_2=\sqrt{v_0^2+2a(x-x_0)}$
$v_2=\sqrt{(0)^2+2(0.9 \ m/s^2)(225 \ m)}=21 \ m/s$
Q#33
(II) A space vehicle accelerates uniformly from 85 m/s at t = 0 to 162 m/s at t = 10.0 s. How far did it move between t = 2.0 s and t = 6.0 s?
Answer:
Calculate the acceleration from the velocity–time data using Eq. $v=v_0+at$ and then use Eq. $x-x_0=v_0t+\frac{1}{2}at^2$ to calculate the displacement at t = 2.0 s and t = 6.0 s. The initial velocity is $v_0$ = 65 m/s.
$a=\frac{v-v_0}{t}=\frac{162 \ m/s - 85 \ m/s}{10.0 \ s}=7.7 \ m/s^2$
$x(2.0 \ s)-0=(85 \ m/s)(2.0 \ s)+\frac{1}{2}(7.7 \ m/s^2)(2.0 \ s)^2=185.4 \ m$ and
$x(6.0 \ s)-0=(85 \ m/s)(6.0 \ s)+\frac{1}{2}(7.7 \ m/s^2)(6.0 \ s)^2=648.6 \ m$, so
$x(6 \ s)-x(2.0 \ s)=648.6 \ m - 185.4 \ m = 463.2 \ m$
Q#34
(III) A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 1.4 $m/s^2$ to his maximum speed of 6.0 m/s which he then maintains.
(a) How long does it take him to catch up to the empty box car?
(b) What is the distance traveled to reach the box car?
Answer:
(a) The train’s constant speed is train v_{min}$$ = 5.0 m/s, and the location of the empty box car as a function of time is given by
$x_{train}$ = $v_{train}t$=(5.0 m/s)t.
The fugitive has $v_0$ = 0 m/s and a = 1.4 $m/s^2$ until his final speed is 6.0 m/s.
The elapsed time during the acceleration is
$t_{acc}=\frac{v-v_0}{a}=\frac{6 \ m/s-0}{1.4 \ m/s^2}=4.286 \ s$
Let the origin be the location of the fugitive when he starts to run. The first possibility to consider is, “Can the fugitive catch the empty box car before he reaches his maximum speed?” During the fugitive’s acceleration, his location as a function of time is given by Eq.
$x-x_0=v_0t+\frac{1}{2}at^2$
$x_{fugitive}=0+0+\frac{1}{2}(1.4 \ m/s^2)t^2$
For him to catch the train, we must have
$x_{train}=x_{fugitive}$
$(5.0 \ m/s)t=(0.7 \ m/s^2)t^2$
The solutions are t = 0 and t = 7.1 s.
Thus the fugitive cannot catch the car during his 4.286 s of acceleration. Now the equation of motion of the fugitive changes. After the 4.286 s of acceleration, he runs with a constant speed of 6.0 m/s.
Thus his location is now given (for times t > 5 s) by the following:
$x_{fugitive}=\frac{1}{2}(1.4 \ m/s^2)(4.286 \ s)^2+(6.0 \ m/s)(t-4.286 \ s)$
$x_{fugitive}=(6.0 \ m/s)t-12.86 \ m$
So now, for the fugitive to catch the train, we again set the locations equal. train fugitive
$x_{train}=x_{fugitive}$
$(5.0 \ m/s)t=(6.0 \ m/s)t-12.86 \ m$
so, t = 12.86 s
(b) The distance traveled to reach the box car is given by the following: fugitive
$x_{fugitive}=(6.0 \ m/s)(12.86 \ s)-12.86 \ m=64 \ m$
Q#35
(III) Mary and Sally are in a foot race (Fig. 2–39). When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.40 $m/s^2$ to the finish line. What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?
Answer:
For the runners to cross the finish line side-by-side, they must both reach the finish line in the same amount of time from their current positions. Take Mary’s current location as the origin. Use Eq.
$x-x_0=v_0t+\frac{1}{2}at^2$
For Sally:
$22 \ m=5.0 \ m +(5.0 \ m/s)t+\frac{1}{2}(-0.40 \ m/s^2)t^2$
$t^2-25t+85 = 0$
$t=\frac{25 \pm \sqrt{25^2-4(1)(85)}}{2}$, so
t = 4.059 s and t = 20.94 s
The first time is the time she first crosses the finish line, so that is the time to be used for the problem. Now find Mary’s acceleration so that she crosses the finish line in that same amount of time.
For Mary:
$22 \ m=0+(4.0 \ m/s)t+\frac{1}{2}at^2$
$22 \ m=0+(4.0 \ m/s)(4.059 \ s)+\frac{1}{2}a(4.059 \ s)^2$
$22 \ m=16.236 \ m+\frac{1}{2}a(4.059 \ s)^2$
$11.528 \ m=a(4.059 \ s)^2$
$a=\frac{11.528 \ m}{(4.059 \ s)^2}=0.70 \ m/s^2$
Q#36
(III) An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 135 km/h. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car’s acceleration is 2.0 $m/s^2$ how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?
Answer:
Given: $v_s=(135 \ km/h)\left(\frac{1000 \ m/km}{3600 \ s/h}\right)=37.5 \ m/s$ and $v_{p1}=(95 \ km/h)\left(\frac{1000 \ m/km}{3600 \ s/h}\right)=26.39 \ m/s$
Define the origin to be the location where the speeder passes the police car. Start a timer at the instant that the speeder passes the police car and find another time that both cars have the same displacement from the origin. For the speeder, traveling with a constant speed, the displacement is given by the following:
$\Delta x_s=v_st=(37.5 \ m/s)t$
For the police car, the displacement is given by two components. The first part is the distance traveled at the initially constant speed during the 1 second of reaction time.
$\Delta x_{p1}=v_{p1}(1.00 \ s)=26.39 \ m$
The second part of the police car displacement is that during the accelerated motion, which lasts for (t − 1.00) s. So this second part of the police car displacement, using Eq. $\Delta x=v_0t+\frac{1}{2}at^2$ is given as follows:
$\Delta x_{p2}=v_{p1}(t-1.00 \ s)+\frac{1}{2}a_p(t-1.00 \ s)^2$
$\Delta x_{p2}=(26.39 \ m/s)(t-1.00 \ s)+\frac{1}{2}(2.60 \ m/s^2)(t-1.00 \ s)^2$
So the total police car displacement is the following:
$\Delta x=\Delta x{p1}+\Delta x_{p2}$
$\Delta x=26.39 \ m + 26.39 \ m (t - 1.00)+(1.30)(t-1.00)^2 \ m$
Now set the two displacements equal and solve for the time.
$37.5t=26.39 \ m 26.39 \ m (t - 1.00)+(1.30)(t-1.00)^2 \ m$
$t^2-10.5t+1=0$
$t=\frac{10.55 \pm \sqrt{(10.55)^2-4.00}}{2}$
t = 0.0957 s and t = 10.5 s
The answer that is approximately 0 s corresponds to the fact that both vehicles had the same displacement of zero when the time was 0. The reason it is not exactly zero is rounding of previous values. The answer of 10.5 s is the time for the police car to overtake the speeder. as a check on the answer, the speeder travels
$\Delta x_s=(37.5 \ m/s)(10.5 \ s)=394 \ m$ and the police car travels
$\Delta x_p=26.39 \ m + 26.39 \ m (10.5 - 1.00)+(1.30)(10.5-1.00)^2 \ m$
$\Delta x_p=394 \ m$
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