Question

Car travelling along a straight road at 14m/s is approaching traffic lights .driver applies brakes and car comes to rest with constant retardation/deceleration. The distance from the point where brakes are applied to where the car comes to rest is 49 m. Find deceleration.

Answer 1

Initial velocity=14m/s

Final velocity=0m/s

Distance travelled=49m

we know,

${v}^{2} = {u}^{2} + 2as$

0=14^2+2*49a

a=-196/98

a=-2m/s/s

Deceleration=2m per second per second.

Answer 2

s=ut+1/2at^2 where s = 49m and u=14 a=? v=? now 49=14×0+ 1/2 at^2 now 49=0+1/2at^2 that is 49/2 =24.5m/s^2