Question

A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1m/s^2, at the same time the man starts running with uniform velocity of 10m/s. What is the minimum time in which the man catches the bus?

Answer 1

$\huge\textbf{ YOUR ANSWER \: }$

Let time taken by man = time taken by bus =x

For bus:

u=0

a=1m/s²

t=x

S=ut+1/2at²  =o × x +1/2 × 1 × x²  =x²/2

For man :

v=10m/s

t=x

S=v×t = 10x

distance travelled by man - distance travelled by bus =48

so, 10x-x²/2 =48

20x-x²/2=48

20x-x²=48×2

20x-x² =96

20x -x²-96 =0

x²-20x+96=0

x²-12x-8x+96=0

x(x-8)-12(x-8)=0

(x-12)(x-8)=0

x=12 or x=8

minimum time = 8sec

$\boxed{\bold{8 sec \:}}$

$\huge\bf{THANKS \:}$