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Calculus/Integration -- acceleration, velocity
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Question
A speeding motorcyclist sees his way blocked by a wagon some distance 's' ahead and slams on his brakes. Given that the brakes impart to the motorcycle a constant negative acceleration 'a' and that the wagon is moving with speed v1 in the same direction as the motorcycle, show that the motorcyclist can avoid collision only if he is traveling at a speed less than v1 + sqrt(2abs(a)s).
Distance: s
Accleration: a (a negative number)
Wagon speed: v (in same direction as motorcycle)
Motorcycle speed: x
For a collision to occur, we would need distance formulas for the motorcycle and wagon.
The wagon is at position s + vt where s, v are constant, t is time.
The motorcycle is at xt - atē/2.
We would want t so that this is always positive, so we need to find what the value of x would be to let the minimum distance between them to be 0 or greater.
Subtract the distance of the wagon from distance of the motorcycle to find out how far in front the wagon is. Take the derivative, and set it equatil to 0 to find where the minimum occurs. This mimimum should always be 0 or greater for there to be no collision.
Thereofre, when the equation is 0, solving for x would give you the maximum speed.
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Answers by Expert:
Scotto
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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.
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