Question

You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s2. You instantly start running toward the still-open door at 5.1 m/s.

A)How long does it take for you to reach the open door and jump in?

B) What is the maximum time you can wait before starting to run and still catch the bus?

Answer 1

(A)

Consider a reference frame, let x=0 is at your initial position. Your position as a function of time is

where v is your velocity, and t=0s is the point when the bus stars to accelerate from rest. Given v=5 m/s, we get

The bus is under a uniformly accelerated motion. Its motion is governed by the kinematics equations. Using kinematics equation

For bus's open door we get

You will catch the bus when your position is the same as the bus's open door position.

Using equation (1) and (2) we get

This is quadratic equation in time. The roots of quadratic equation are

We have

You reach the open door first at 2.35 seconds. If you still continue your run at 5m/s you will leave the open door behind you. The accelerating bus will eventually pass you again at 7.65 seconds if you had continued your run. Therefore, the time in which you will reach the open door and jump in is 2.35 seconds.

(B) If you wait for time before chasing the door, your position in terms of time when you start to run ( ) is

The bus position is still given by equation (2), you will catch the door when

For minimum we require

Substituting into equation (3) we get