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Solved: Catching the Bus. A student is running at her top
Chapter 2, Problem 97CP(choose chapter or problem)
Problem 97CP
Catching the Bus. 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 0.170 m/s2. (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 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 their specified motions. Explain the significance of this second solution. How fast is the bus traveling 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 have to just catch up with the bus? For what time and what distance does she have to run in that case?
Questions & Answers
QUESTION:
Problem 97CP
Catching the Bus. 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 0.170 m/s2. (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 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 their specified motions. Explain the significance of this second solution. How fast is the bus traveling 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 have to just catch up with the bus? For what time and what distance does she have to run in that case?
ANSWER:
Solution to 97CP
Step 1
Initial velocity of the student=Vg=5m/s
Initial velocity of bus =Vb=0m/s
Initial position of student =0m
Initial position of bus =40m
Initial acceleration of student=0m/s2
Initial acceleration of bus=0.170m/s2
(a)
Here to overtake the bus the position of the bus and the student’s should be same.
..............................................(1)
.................................(2)
Equating (1) and (2)
.......................(3)
Solving this quadratic equation we get two values of t
t=49.27s
t=9.55s
Take t=9.55s
Thus xg=5x9.55=47.75m