Question

You have parked a Lloyd Noble a little late for your classes, so you need to catch the next bus if possible. Running at your top speed from behind the bus, the bus starts accelerating when you are still some distance behind the door of the bus. Assume the bus accelerates at a constant rate. The figure to the right shows you and the bus just as the bus starts moving. It also defines the positive x-direction, the point to use as 푥ൌ0 (the position of the door of the bus when it first starts to move), and the distance, d, from you to the door at this initial time. The fundamental question is: are you going to be able to catch the bus?

You have parked a Lloyd Noble a little late for your classes, so you need to catch the next bus if possible. Running at your top speed from behind the bus, the bus starts accelerating when you are still some distance behind the door of the bus. Assume the bus accelerates at a constant rate. The figure to the right shows you and the bus just as the bus starts moving. It also defines the positive x-direction, the point to use as x 0 (the position of the door of the bus when it first starts to move), and the distance, d, from you to the door at this initial time. The fundamental question is: are you going to be able to catch the bus?

Answer 1

Solution:

Let the acceleration of the bus = a

Let t be the time taken by the person to catch the bus and v be his top velocity .

Distance traveled by the bus in the time t , starting from rest = 1/2 a t².

Distance to be traveled by the person to catch bus = d + 1/2 a t ²

In order to catch the bus, the distance traveled by the person  must equal the distance traveled by the bus in that same time t and the distance d.

vt = d + 1/2 at²

v = d/t + 1/2 a t should be the top speed in order to catch the bus.

you will be able to catch the bus at that top speed.