Question

The breaks applied to the car produce and acceleration of 6m/s^2 in opposite direction to the motion. if the car takes 2s to stop after application of breaks. calculate the distance it travels during this time.

Answer 1

$<b>Answer:$

acceleration (a) = 6m/s²

time = 2secs

final velocity (v) = 0

initial velocity ( u) = ?

Distance (s) = ?

Using first equation of motion :

v = u + at

u = v - at

=> u = 0 - (-6) × 2

=> u = 12m/s

Now, using third equation of motion

s = ut + 1/2 at²

=> s = 12 × 2 + 1/2 ×(- 6 )× 2²

=> s = 24 - 12

=> s = 12 m

Therefore, the car travels 12 m.

Thanks for the question!

Answer 2

Answer:

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.