Question

The brakes applied to a car produces an aceleration of 6m/s square in the opposite direction to the motoin. If the car takes two seconds to stop after the application of brakes then calculate the distance it travels during this time

Answer 1

Answer:

Acceleration a = -6 m/s^2

Time t = 2 s

Final velocity v = 0 m/s

Let initial velocity be u

Let distance be s

v = u + at

So, 0 = u + (-6)(2)

So, u = 12 m/s

Now, s = ut + (1/2) at^2

So, s = 12(2) + (1/2)(-6)(2^2)

So, s = 24 - 12

So, s = 12 m

Thus, distance travelled is 12 m

Answer 2

Answer:

Hey mate....

Here's ur answer :::::::::::::""""

Given : Acceleration (a) = –6m/s²

time (t) = 2s

final velocity (v) = 0m/s

We know that ,

v = u + at

0 = u + (–6) × 2

.°. u = 12 m/s

Also,

s = ut + 1/2 at²

.•. s = 12 * 2 + 0.5 * (–6) * 2²

, s = 24 – 12

.°. s = 12 m.

Thus, the car will move 12 m before it stops after the application of brakes.

Hope it helps!!!

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