Asked by Mike

I need help with this problem:
A car traveling at 43 ft/sec decelerates at a constant 5 ft per second squared. How many feet does the car travel before coming to a complete stop?

I'm not sure where to start, i know how to find antiDerivatives, but i don't even know the equation...


V = dx/dt = 43 - 5 t
Integral of dx = distance travelled = X
X = Integral of Vo - at = (43 - 5t) dt from t = 0 until the time it takes to stop.

The time it takes to stop is Vo/a = 43/5 = 8.6 s

X = Vo^2/a - (1/2)*a*(Vo/a)^2
X = (1/2) Vo^2/a

This formula is often used in physics, but here I have assumed you wanted a mathematical derivation

thanks! perfect!

i need the anti derivative of 2x/x^2 thanks.
Answered by Dillon
The car is travelling 43ft/sec and decelerating at 5ft/s so we can assume it takes 43/5 seconds to stop. The car is slowing down constantly so the average speed of the car can be determined by 43/2 and then multiplying by the time it takes to stop the car we can find how many feet the car traveled before it stopped. (43/2 times 43/5 = 184.9 ft.)
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