Question

A particle is pushed along a horizontal surface in such a way that it start with a velocity of 12 metre per second. Its velocity decreases at a rate of 0.5 metre per second square. find the distance covered by it before coming to rest

Answer 1

Answer:

144 m

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

Answer:

The particle covers 144m before stopping.

Explanation:

${v}^{2} = {u}^{2} + 2as$

$0 = (12ms)^{2} + 2( - 0.5ms)^{2}$

$(1ms)^{2} = 144 {m}^{2} {s}^{2}$

$s = \frac{144 {m}^{2} {s}^{2} }{1m {s}^{2} } = 144m$

So, particle covers 144m before stopping