Question

The brakes applied to a car produce an acceleration of 6ms-² in the opposite direction motion. If the car takes 2 seconds to stop after the application of the brakes, calculate the distance traveled during this time.

Answer 1

Answer:

acceleration =6ms^2

time=2s

final velocity v=0m/s

Let initial velocity be u

Let distance be s

v=u+at

so,0=u+(-6)(2)

so,u=12m/s

Now,s=ut+(1/2)at^2

so,s=12(2)+(1/2)(-6)(2^2)

so,s=24-12

so,s=12m

thus distance travelled is 12m

Answer 2

Given:-

→Acceleration of the car = 6m/s²

(in the opposite direction of motion)

→Time taken by the car to stop = 2s

To find:-

→Distance travelled by the car in given time

Solution:-

In this case:-

•Acceleration of the car will be -6m/s² as it is in the opposite direction of motion.

•Final velocity of the car will be zero as the car finally comes to rest after application of brakes.

By using the 1st equation of motion,we get:-

=>v=u+at

=>0=u+(-6)2

=>-u = -12

=>u = 12m/s

Hence,we have got initial velocity of the car as 12m/s.

Now,by using the 2nd equation of motion,we get:-

=>s=ut+1/2at²

=>s=12(2)+1/2(-6)(2)²

=>s=24+(-3)4

=>s=24-12

=>s=12m

Thus,distance travelled by the car is 12m.