Question

A truck running at 72 km h, is brought to rest
over a distance of 25 m. Calculate the retardation
and the time for which brakes are applied.​

Answer 1

Answer:

Initial speed = 72 km/h = 72×(5/18) = 20 m/s.

to get reatardtaion, equation to be used is " v2 = u2 - 2×a×S " ,

where v is final speed ( v=0 here ), u is initial speed, a is retardation and S is distance travelled

hence retardation a = (20×20) / (2×25) = 8 m/s2

time t is obtained using equation " v = u - a×t ", with v = 0

hence time t = 20/8 = 2.5 s

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Initial speed = 72 km/h = 72×(5/18) = 20 m/s.

to get reatardtaion, equation to be used is " v2 = u2 - 2×a×S " ,

where v is final speed ( v=0 here ), u is initial speed, a is retardation and S is distance travelled

hence retardation a = (20×20) / (2×25) = 8 m/s2

time t is obtained using equation " v = u - a×t ", with v = 0

hence time t = 20/8 = 2.5 s