Question

A motor bike running at 90km/h is slowed to 54km/h by the application of brakes over a distance of 40 m if the brakes are applied with the same force calculate the :

1 total time in which bike comes to rest

2 total distance travelled by the bike

Answer 1

Motor bike changed the speed from 90 km/h to 54 km/h in a distance 40m.

To find the retardation, equation to be applied :- " v2 = u2- 2×a×S " ,
where u and v are initial and final speed respectively, a is retardation and S is ditance travelled

Initial speed = 90 km/h = 90 ×(5/18) = 25 m/s ; Final speed 50 km/h = 50×(5/18) = 15 m/s

hence retardation a = [ (25×25) - (14×14) ] / (2×40) = 5 m/s

(1) time for bike to come to rest :- equation to be applied " v = u - a×t " with v = 0

hence time t to come to rest = u/a = 25/5 = 5 s

(2) distance S covered by bike after applying break :- final speed v =0, hence S = u2 /(2a)

S = (25×25)/(2×5) = 62.5m