A student is running at her top speed of 5.2m/s to catch a bus, which is stopped

Question

A student is running at her top speed of 5.2m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 40.1m from the bus, it starts to pull away, moving with a constant acceleration of 0.164m/s^2 . Part A For how much time does the student have to run at 5.2m/s before she overtakes the bus? ANSWER= 8.98s Part B For what distance does the student have to run at 5.2m/s before she overtakes the bus? ANSWER=46.7m Part C When she reaches the bus, how fast is the bus traveling? ANSWER=1.47m/s Part D If the student's top speed is 2.90m/s , will she catch the bus? ANSWER= NO Part E What is the minimum speed the student must have to just catch up with the bus? Part F For what time does she have to run in that case? Part G For what distance does she have to run in that case? I just need part E/F/G.. PLEASE HELP will rate 5 stars. Thanks

Explanation / Answer

E. let the velocity be v in time the distance travelled by bus is given by 1/at^2 because initial velocity is zero

now vt = 40.1+1/2at^2

or, v = 40.1/t +1/2at

as velocity of the girl is constant dv/dt = 0

or, 0 = -40.1/t^2 + at

solving we get t = 6.3s

now v = 6.88 m/s

F. she has to run for 6.3s

G. the distance travelled by the girl is 40.35 m

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