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Help please!!! A car braked with a constant deceleration of 40 ft/s^2. What is the...

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user3498849 | (Level 2) eNoter

Posted May 13, 2013 at 7:11 AM via web

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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!!

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rakesh05 | High School Teacher | (Level 1) Assistant Educator

Posted May 13, 2013 at 7:28 AM (Answer #1)

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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.

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