Question

A motor bike running at 90kmh-1

, is slowed down to 54kmh-1

by the

application of brakes, over a distance of 40m. If the brakes are applied

with the same force, calculate (i) total time in which bike comes to

rest (ii) total distance travelled by bike.​

Answer 1

Answer:

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 distance 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

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