Question

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

Answer 1

v=0
t=2
a=6
a=v-u/t
a=o-u/2
6=-u/2
6*2=u
12=u
distance(s)=2as=v2-u2
2(6)s=(0)2-(12)2
12s=-144
s=144/12
s =12

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.