Question

a man is 48m behind a bus which is at rest. the bus starts accelerating at the rate of 1m/s2; at the same time the man starts running with uniform velocity of 10m/s. what is the minimum timein which the man catches 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
minimum time = 8sec

Answer 2

Explanation:

The minimum time in which the man catches the bus is 8 sec....

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