Question

A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12 m/s, and decreases at the rate of 0.5 m/s2. The time it will take to come to rest is:
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Answer 1

Answer:

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Explanation:

U = 12 m/s

a = - 0.50 m/s/s

v = 0

v = u + a t

=> t = (0 - 12 ) / (-0.50) = 24 sec

s = u t + 1/2 a t² = 12 * 24 - 1/2 * 0.50 * 24²

= 144 meters

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Answer 2

rate of retardation = 0.5m/s2

initial velocity(u)= 12m/s

final velocity (v)= 0 m/s

since

acceleration= (v-u)/t

=> 0-12/t = -0.5 ( -0.5 because retardation is negative acceleration

=> -12/t = -0.5

=> t= 12/0.5

therefore , time = 24s