A motorbikie running at 90 km/h is slowed down to 50 km/h by application of brakes over a distance of 40m.If the brakes are applied with the same force.Calculate : (1) total time in which bike comes to rest ? (2) total distance covered by the bike ?
Asked by nishasaumitya | 22nd May, 2018, 11:56: PM
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.
Answered by Thiyagarajan K | 23rd May, 2018, 12:23: PM
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