Question

Two breaks are applied to a car produce an acceleration of 6 m/s in opposite direction to motion. If car takes 2s to stop after applications of breaks, calculate distance.

Answer 1

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Step-by-step explanation:

Given,

Acceleration, a = -6 m/s²

Time taken,  t = 2 s

Final velocity, v = 0 m/s (As it stops.)

To Find,

Distance travelled, s = ?

Formula to be used,

1st equation of motion, v = u + at

2nd equation of motion, s = ut + 1/2 at²

Solution,

Putting all the values, we get

v = u + at

⇒ 0 = u + (- 6) × 2

⇒ u = 12 m/s

Here, the initial velocity is 12 m/s.

Now, the distance travelled,

s = ut + 1/2 at²

⇒ s = 12 × 2 + 1/2 × (- 6) × 2²

⇒ s = 24 - 12

⇒ s = 12 m

Hence,the distance travelled is 12 m.