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/s2.

A. For how much time and what distance does the student have? 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

This question contains many parts which can be solved one by one but here in all parts we use the concept of velocity, speed and acceleration. They all are interrelated with each other as all are related to motion in one dimension.

Complete step by step answer:
For convenience, let the students' (constant) speed be ${v_0}$ and the bus's 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
${x_1} = {v_0}t$ and ${x_2} = {x_0} + \dfrac{1}{2}a{t^2}$
A. Putting ${x_1} = {x_2}$ and solving for the time t gives
$t = \dfrac{1}{a}\left( {{v_0} \pm \sqrt {v_0^2 - 2a{x_0}} } \right)$
Substitute the known values in this equation.
$\begin{array}{c} t = \dfrac{1}{{0.2}}\left( {5 \pm \sqrt {{5^2} - 2 \times 0.2 \times 40} } \right)\\\ = 10.40s \end{array}$
Therefore, the distance run by the student is
$\begin{array}{c} D = {v_0}t\\\ = 5\;m/s \times 10\;s\\\ = 50\;m \end{array}$
B. The speed of the bus is

$\begin{array}{c} v = at\\\ = 0.2\;m/{s^2} \times 10\;s\\\ = 2\;m/s \end{array}$

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 bus's velocity is
$\begin{array}{c} v = 0.2\;m/{s^2} \times 40\;s\\\ = 4\;m/s \end{array}$

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 hence the minimum speed is $\sqrt {2 \times 0.2 \times 40} = 40\;m/s$
She will run for the time $\dfrac{{4\;m/s}}{{0.2\;m/{s^2}}} = 20\;s$ and it covers the distance of $4\;m/s \times 20\;s = 80\;m$
Note: In Physics, when the position of an object changes over a period of time is known as motion. Mathematically the motion is described in terms of displacement, distance, velocity, speed, acceleration, and time. By attaching the frame of reference, the motion of a body is observed. Further, based on the change in position of the body relative to the frame, the motion is measured.