Question

The breaks applied on a car produce an acceleration of 6m/s2 in the opposite direction to motion. If the car takes 2sec.to stop after application of breaks calculate the distance it travels this time.

Answer 1

Acceleration (a) = -6m/s^2(as it is acting in the opposite direction)
Time taken (t) = 2s
Final velocity (v) = 0 (as it comes to rest)

Let initial velocity be u
Using the first equation of motion
v = u +at
0 = u +(-12)
u = 12
So, the initial velocity was 12m/s

Let distance travelled be s
Using 3rd equation of motion

2as= v^2 -u^2
2 (-6)(s) =(0)^2 -(12)^2
-12s = -144
s =12
So, the distance travelled by the body was 12m.

Answer 2

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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.