Question

a passenger is at a distance of x from a bus when the bus begins to move with constant acceleration a. what is the minimum velocity with which the passenger should run towards the bus to catch it?

A) (2ax)^1/2

B) 2ax

C) (ax)^1/2

D) ax

Answer 1

Lets assume that the man will catch the bus after time t and the passenger's velocity is v.
So,
$$\begin{gathered} Dis\tan cecoverdbythepassenger=Dis\tan cetravelledbythebus+x \\ \Rightarrow vt=\frac{1}{2}a{t}^2+x \end{gathered}$$
On the other hand,
$$\begin{gathered} velocityofthepassenger=velocityofthebus \\ \Rightarrow v=at \\ \Rightarrow t=\frac{v}{a} \end{gathered}$$
So,
$$\begin{gathered} v\frac{v}{a}=\frac{1}{2}a\frac{v^2}{a^2}+x \\ \Rightarrow \frac{1}{2}\frac{v^2}{a}=x \\ \Rightarrow v=\sqrt{2ax} \end{gathered}$$
Minimum velocity of the passenger is $\sqrt{2ax}$.
A) is the right choice.