# 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!!

### 1 Answer | Add Yours

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.