A student is running at her top speed of $5.0\,\text{m/s}$ to catch a bus, which is stopped at the bus stop. When the student is still $40.0\,\text{m}$ from the bus, it starts to pull away, moving with a constant acceleration of $0.170\,\text{m/s}^2$.

Question

A student is running at her top speed of $5.0\,\text{m/s}$ to catch a bus, which is stopped at the bus stop. When the student is still $40.0\,\text{m}$ from the bus, it starts to pull away, moving with a constant acceleration of $0.170\,\text{m/s}^2$.(a) For how much time and what distance does the student have to run at $5.0\,\text{m/s}$ before she overtakes the bus?
, (b) When she reaches the bus, how fast is the bus traveling?
, (c) Sketch a position-time graph for both the student and the bus. Take $x = 0$ at the initial position of the student.
, (d) The equations used in part (a) have a second solution for time, 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 traveling at this point?
, (e) If the student&#8217;s top speed is $3.5\,\text{m/s}$, will she catch the bus?
, (f) What is the minimum speed the student must have to just catch up with the bus? For what time and distance does she have to run in that case?
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Jason · Accepted Answer

(a) The student must run for approximately $ 9.55 \, \text{s} $ and cover $ 47.77 \, \text{m} $.
(b) When she reaches the bus, it is traveling at $ 1.62 \, \text{m/s} $.
(c) **Graph Description**:
&#8211; Student&#8217;s Path: A straight line with constant slope at $ 5.0 \, \text{m/s} $.
&#8211; Bus&#8217;s Path: A parabola starting at $ 40.0 \, \text{m} $ with increasing slope.
(d) Second solution $ t \approx 49.27 \, \text{s} $ represents when the bus overtakes the student again. Bus speed at that time: $ 8.38 \, \text{m/s} $.
(e) If the student&#8217;s top speed is $ 3.5 \, \text{m/s} $, she will not catch the bus (no real solution, $ D < 0 $).
(f) Minimum speed to catch the bus: $ 3.69 \, \text{m/s} $, time $ 21.72 \, \text{s} $, covering $ 80.17 \, \text{m} $.

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