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

Answer:

Part a)

$t = 16.8 s$

$d = 100.8 m$

Part b)

$v_f = 2.86 m/s$

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

$x_{bus} = x_{student}$

$0 + \frac{1}{2}at^2 + d = v_{student} t$

$\frac{1}{2}(0.170)t^2 + 60 = 5 t$

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

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

$v_f = v_i + at$

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

$v_f = 2.86 m/s$