Question

the brakes are applied to a car produce on acceleration of 6m/s in the opposite direction of the motion . If the car takes 2s to stop after the application of brakes . Calculate the distance traveled by the car during the time

Answer 1

Given that the acceleration i

was produced in the opposite direction of the motion. So it is retardation.

a = -6m/s²

Final velocity of the car = 0

Time = t = 2s

So finding initial velocity of the car by the relation ,

v = u + at

0 = u +(-6)2 = u-12

u = 12 m/s

Then, finding the distance covered during that time , by using the relation : .

s = ut + ½ at²

= 12× 2+½+(-6)×2² = 24 - 12

= 12 m

The required distance = 12 m

Answer 2

GIVEN::

Acceleration= -6m/s² [Since, it is retardation]

Final velocity (v) = 0 [Since, the car stops finally.]

Time taken = 2 sec

Initial velocity(u) = ?

Distance travelled(s) = ?

LETS FIND INITIAL VELOCITY (u)

To find velocity we have to apply first equation of motion i.e. v=u + at

⇒v = u + at

⇒0 = u + (-6 × 2)

⇒0 = u - 12

⇒-u = -12

⇒u = 12.

Therefore, initial velocity is 12m/s.

LETS FIND THE DISTANCE TRAVELLED (s)

To find the distance travelled we have to apply 2nd equation of motion.

⇒s = ut + 1/2at²

⇒s = (12 × 2) + (1/2 × -6 × 2²)

⇒s = 24 + (-3 × 4)

⇒s = 24 - 12

⇒s = 12m.

Therfore, the distance travelled by the car is 12m.