Question

braker applied to a car produce an acceleration of 6m/s^2 in opposite direction to motion if the car takes 2 second to stop after application of brakes calculate the distance

Answer 1

We have,
u = ?
v = 0
a = -6 m/s²
t = 2 seconds
s(distance) = ?

We know that,

v = u+at

In case of deceleration or negative acceleration, formula becomes,

v = u-at

Now, Putting values in above formula,
0 = u-6×2
-u = -12
[u = 12 m/s]•••••••••••(1)

By using 3rd equation of uniformly accelerated motion,
that is,

v²=u²+2as

In case of deceleration or negative acceleration, formula becomes,

${v}^{2} = {u}^{2} - 2as \\Now\: Putting\:values\: in\: above\: formula, \\ {0}^{2} = {12}^{2} - 2 \times 6 \times s \\ - {12}^{2} = - 12s \\ 12s = 12 \times 12 \\ s = \frac{12 \times 12}{12} \\ s = 12 \: metres$

Distance covered in 2 seconds is 12 metres.

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