Question

A student is running to catch the campus shuttle bus, which is stopped at the bus stop. The student is runnign at a constant speed of 6.0 m/s; she cant run any faster. When trhe student is still 60 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170m/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?

Answer 1

Part a)

`t = 16.8 s`

`d = 100.8 m`

Part b)

`v_f = 2.86 m/s`

Step-by-step explanation:

Part a)

Constant speed by which the student will run is given as

`v = 5 m/s`

now after some time if student is going to overtake the position of bus

so here the final positions will be same

so we have

`x_(bus) = x_(student)`

`0 + (1)/(2)at^2 + d = v_(student) t`

`(1)/(2)(0.170)t^2 + 60 = 5 t`

`0.085 t^2 - 5t + 60 = 0`

so it is

`t = 16.8 s`

So student will run the total distance

`d = vt`

`d = (6)(16.8)`

`d = 100.8 m`

Part b)

Speed of bus when student reach the bus is given as

`v_f = v_i + at`

`v_f = 0 + (0.170)(16.8)`

`v_f = 2.86 m/s`