Chapter 2: Q. 60 (page 62)

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 4.5 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?

Short Answer

Expert verified

a. It will take $3 s$ to reach the door and jump up.

b. The maximum time you can wait before starting to run

and still catch the bus is 3.5 s.

Step by step solution

Part a; Step 1: Given data

Distance of the person from the bus, $s = 9.0 m$

The bus has an acceleration of $a = 1 m / s^{2}$

Initial speed of the person, $u = 4.5 m / s$

Determination of time

The relative acceleration between the bus and the person is, $a' = - 1 m / s^{2}$

Now, according to the question, the distance between the door and the person is given by,

$s = u t + \frac{1}{2} a t^{2} 9 = 4.5 t - \frac{1}{2} t^{2} 0.5 t^{2} - 4.5 t + 9 = 0 t^{2} - 9 t + 18 = 0 t^{2} - 6 t - 3 t + 18 = 0 (t - 3) (t - 6) = 0 t = 3 s, t = 6 s$

It will take $3 s$ by the person to reach the bus and jump in.

Part b; Step 1: Introduction

Distance travelled by the person can be written as,

$x = v t x = 4.5 t$

Let $t'$ be the time the person has to wait.

$d'$ be the relative distance

Calculation of the time that the person can wait for

$4.5 t = d' + u t + 0.5 a t^{2} 4.5 t = (9 + 0.5 a t'^{2}) + a t' + 0.5 a t^{2} 4.5 t = 9 + 0.5 t'^{2} + t' t + 0.5 t^{2} 0.5 t^{2} - (4.5 - t') t + (9 + 0.5 t'^{2}) = 0$