Question

an engine is moving with a velocity 44m/s after applying brakes it stops after covering a distance of 121 m. calculate retardation and time taken by the engine to stop.

Answer 1

1936=2@121
@=968/121=8
retardation is 8 m/s²
44=8t
t=5.5s
time is 5.5s

Answer 2

Answer:I will give you all a clear explanation......

Explanation: We know that V^2=U^2+2as

Now V=0,U=44m/s,S=121 m

Hence 0^2=44^2+2×a×121

0=1936+2×a×121

-1936=2×a×121

-1936/2=121×a

-968=121a

-968/121=a

Therefore a=-8 (so it is the retardation or we can also write retardation =-8)

Now if a=-8,then

V=U+at

So, 0=44-8×t

-44=-8×t

44=8×t (minus and minus cancels)

44/8=t

5.5=t

Therefore t=5.5 s

Therefore retardation is 8 m/s^2 and time is 5.5 seconds.

Hope it is clear.