An engine is moving with a velocity 44 m/s. After applying the brakes, it stops after covering a distance of 121 m. Calculate retardation and time taken by the engine to stop.

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Solution

Here,

u = 44 m/s , S =121 m , v = 0

Using , $v^{2}$ = $u^{2}$ + 2aS

0 = $44^{2}$ + (2 x 121 x a)

a = -8 $m / s^{2}$

Also,

v = u + at

0 = 44 - 8t

t = 5.5s

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An engine is moving with a velocity 44 m/s. After applying the brakes, it stops after covering a distance of 121 m. Calculate retardation and time taken by the engine to stop.

Here, u = 44 m/s , S =121 m , v = 0 Using , v2 = u2 + 2aS 0 = 442 + (2 x 121 x a) a = -8 m/s2 Also, v = u + at 0 = 44 - 8t t = 5.5s

Here,

u = 44 m/s , S =121 m , v = 0

Using , $v^{2}$ = $u^{2}$ + 2aS

0 = $44^{2}$ + (2 x 121 x a)

a = -8 $m / s^{2}$

Also,

v = u + at

0 = 44 - 8t

t = 5.5s