Help please!!! A car braked with a constant deceleration of 40 ft/s^2. What is the distance covered before the car comes to a stop? Thank you sooooo much!!

Expert Answers
rakesh05 eNotes educator| Certified Educator

Let the car is moving with velocity u ft/second. And final velocity of the car is v ft/second. After appllying the break the deceleration f of the car is given by  f=-40 ft/second^2.

As when the car comes to a stop,   v=0.

      Now we know that    v^2=u^2+2fs.

Where s is the distance travelled by the car.

So,                               0^2=u^2+2(-40)s

or,                                  0=u^2-80s

or,                                 80s=u^2

or,                                   s=u^2/80 ft.

Means after moving by   u^2/80  ft. the car comes to a stop.