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.
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1 answer

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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answered by Dillon
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