Question

A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away moving with a constant acceleration of 0.2 $m{s}^{-2}$.
a.For how much time and what distance does the student? have to run at 5.0 m/s before she overtakes the bus?
b.When she reached the bus, how fast was the bus travelling?
c.Sketch an x-t graph for both the student and the bus.
d.The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and the bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus travelling at this point?
e. If the student's top speed is 3.5 m/s, will she catch the bus?
f. What is the minimum speed the student must catch up with the bus? For what does she have to run in that case?

Answer 1

For convenience, let the students (constant) speed be $v_0$ and the buss initial position be $x_0$. Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero and the initial velocity of the bus is taken to be zero. The positions of student $x_1$ and bus $x_2$ as functions of time are then
$x_1=v_{0}t,x_2=x_0+\left(l/2\right)a{t}^2$
A. Setting $x_1=x_2$ and solving for the time t gives
$t=\left(1/a\right)(v_0\pm \sqrt{v_0^2-2a{x}_0})$
= $\left(1/0.2\right)\left(\right(5.0)\pm \sqrt{5^2-2\left(0.2\right)\left(40.0\right)})$ = 10.40 s
The student will be likely to hop on the bus the first time she passes it [see part (d) for a discussion of the later time]. During this time, the student has run a distance $v_0$t = (5 m/s)(10 s) = 50 m

B.The speed of the bus is v = at = 0.2 x 10 = 20 m/s

C.The results can be verified by noting that the x lines for the student and the bus intersect at two points.

D. At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up with her. At this later time, the buss velocity is (0.2 $m{s}^{-2}$) (40 s) = 8.0 m/s.

E. No; $v_0^2<2a{x}_0$, and the roots of the quadratic are imaginary.When the student runs at 3.5 m/s the two lines do not intersect.

F. For the student to catch the bus, $v_0^2<2a{x}_0$ and so the minimum speed is $\sqrt{2\left(0.2m{s}^{-2}\right(40m)}$ = 4.0 m/s. She would be running for a time $\frac{4.0m/s}{0.2m{s}^{-2}}$ = 20s, and cover a distance of (4.0 m/s)(20 s) = 80.0 m.
However, when the student runs at 4.0 m/s, the lines intersect at one point (x = 80 m).