Question

A bus starts to move with acceleration 1m/s^2. A man who is 48 m behind the bus runs to catch it with a constant velocity of 10 m/sec. In how much time he will catch the bus?

Answer 1

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

=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