Question

A body is moving with a velocity of 12m/s and it comes to rest in 18m, what was the acceleration? a = -6 m s -2 a = 4 m s -2 a = -4 m s -2 a = 2 m s -2

Answer 1

SOLUTION :-

$velocity \: (u) = 12 \frac{m}{s} \\ \\ distance \: (d) = 18m \\ \\ final \: velocity \: (v) = 0 \: \\ \\ \: from \: \: second \: equation \: of \: motion \: \\ \\ {v}^{2} - {u}^{2} = 2( - a)s \\ \\ = > {0}^{2} - {12}^{2} = 2( - a)(18) \\ \\ = > - 144 = - 36a \\ \\ = > a = \frac{ - 144}{ - 36} = 4 \: \frac{m}{ {s}^{2} }$

∴ ACCELERATION = 4 m/s²

[ NOTE : HERE ACCELERATION IS NEGATIVE BECAUSE IT IS IN THE OPPOSITE DIRCTION OF MOTION . AND FINAL VELOCITY IS 0 SINCE IT CAME TO REST ]

Answer 2

Answer:

-4ms^-2

Explanation:

v^2 = u^2 + 2aS

Since v=0, v^2=0

-u^2 = 2aS

a = -u^2/2S = -(12)^2 / 2(18) = -144/36 = -4ms^-2