Question

A man is 45 m behind the bus when the bus start accelerating from rest with acceleration $2.5$. With what minimum velocity should the man start running to catch the bus

• A. $12$
• B. $14$
• C. $15$
• D. $16$

Answer 1

The correct option is C $15$

Given,

$u=0m/s$ Speed of bus

$a=2.5m/s^2$

From 2nd equation of motion,

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

$s=0+\frac{1}{2}\times 2.5t^2$

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

Lets consider, $u=$ initial velocity of man

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

$ut=45+s$

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

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

For minimum velocity,

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

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

$45=1.25t^2$

$1.25t^2=45$

$t^2=\frac{45}{1.25}=36$

$t=\sqrt{36}s$

$t=6s$

From equation (2),

$u_{min}=\frac{45}{6}+1.25\times 6$

$u_{min}=7.5+7.5$

$u_{min}=15m/s$

The correct option is C.