Question

A man is $45m$ behind the bus when the bus starts accelerating from rest with acceleration of $2.5m{s}^{-2}$. With what minimum velocity should the man start running to catch the bus?

• A. $12m{s}^{-1}$
• B. $14m{s}^{-1}$
• C. $15m{s}^{-1}$
• D. $16m{s}^{-1}$

Answer 1

The correct option is C

$15m{s}^{-1}$

Step 1: Given Data:

Distance, $s=45m$

Acceleration of bus, $a=2.5m{s}^{-2}$

$u=0m/s$ speed of bus

Step 2: Formula Used:

$s=ut+\frac{1}{2}a{t}^2$

Where, $u=$initial speed

$v=$Final speed

$a=$acceleration

$t=$time

Step 3: Calculating the required speed:

$s=0+\frac{1}{2}\times 2.5\times t^2$ [Where, ithe nitial velocity is zero as bus starts from rest.]

$\Rightarrow$$s=\frac{2.5}{2}{t}^2$….(1)

From the above figure, man needs to cover distance $(45+s)m$ at time $t$ to catch the bus

$ut=45+s$

$\Rightarrow$$ut=45+\frac{2.5}{2}{t}^2$ [from(1)]

$\Rightarrow$ $u=\frac{45}{t}+\frac{2.5}{2}t$ ….(2)

For minimum velocity,

$\frac{du}{dt}=0$

$\Rightarrow$$-\frac{45}{t^2}+1.25=0$

$\Rightarrow$ $45=1.25t^2$

$\Rightarrow$ $\begin{array}{rcl}{t}^2& =& \frac{45}{1.25}\\ & =& 36\end{array}$

$\Rightarrow$ $t=\sqrt{36}s$

$\Rightarrow$ $t=6s$

From equation(2),

$\begin{array}{rcl}{u}_{min}& =& \frac{45}{6}+1.25\times 6\\ & =& 7.5+7.5\\ & =& 15m/s\end{array}$

Hence, option C is the correct answer.