Determing Antiderivatives from a Word Problem

"A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/ . What is the distance covered before the car comes to a stop?"

I know that the antiderivative of the acceleration is the velocity, and the antiderivative of the velocity is its distance, but I need help on how to derive the equations from this word problem.

Re: Determing Antiderivatives from a Word Problem

I would have written it slightly differently but basically the same work as Aryth did:

"A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/ s^{2}. What is the distance covered before the car comes to a stop?"

One thing you have to be careful about is that speed is give in "miles per hour" while acceleration is given in "feet per second per second". We must change to the same units. There are 5280 feet per mile so 50 miles per hour is 50(5280)= 264000 feet per hour. There are 60 minutes per hour and 60 seconds per minute so there are 3600 seconds per hour. 264000 feet per hour is 26400/3600= 73 and 1/3 feet per second.

Acceleration is the derivative (rate of change) of velocity so (negative because this is **de**celeration).

Integrating, v= -22t+ C and since it had an initial (t= 0) speed of 73 and 1/3 feet per second, v= -22t+ 73 1/3.

v= dx/dt= -22t+ 73 1/3 so . Since we want to know how far the car travels before it stops, we can take x= 0 as the initial position and so C= 0. The car will have stopped when v= 0 so solve -22t+ 73 1/3= 0 for t and put that value of t into .