Question

A motor bike running at 90 km/h to 50 km/h by the application of brakes, over a diatance of 40 m. if the brakes are applied with the same force, calculate (i) total time in which bike comes to rest (ii) total distance travelled by the bike​

Answer 1

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Explanation:

Answer 2

Answer:

58 m

Explanation:

Motor bike changed the speed from 90 km/h to 50 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) = 13.88 m/s ≈ 14 m/s

 

hence retardation a = [ (25×25) - (14×14) ] / (2×40)  = 5.36 m/s2

 

(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  = 25/5.36  = 4.7 s

 

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

 

S = (25×25)/(2×5.36) = 58 m.