Question

A student is running at her top speed of 4.9 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 38.1 m from the bus, it starts to pull away, moving with a constant acceleration of 0.175 m/s2.a) For how much time and what distance does the student have to run at 5.0m/s before she overtakes the bus?

b) When she reaches the bus how fast is the bus traveling

c) Sketch an x-t graph for both the student and the bus. Take x=0 at the initial position of the student

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 bus are again at the same place if they continue thier specified motion. Explain the significance of this second solution. How fast is the bus traveling at this point

e) If the students top speed is 3.5 m/s will she catch the bus?

f) is the minumun speed the student must have to just catch up with the bus? For what time and distance must she run in that case

Answer 1

(a) Time required for her to catch up with the bus is 9.05seconds

(b) The speed of the bus at that time is 1.58 m/s

(c) see attachment below

(d) See the attachment below

(e) No, she will not catch up with the bus.

Step-by-step explanation:

See attachment below for the full solution to the problem