asst9-f10.solns

asst9-f10.solns - MATH 137 Assignment 9, Part 2 Fall 2010...

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MATH 137 Assignment 9, Part 2 Fall 2010 1. A car is traveling at 60 km/h on a straight road when the brakes are applied, producing a constant deceleration of 8 m/s 2 . What is the distance traveled before the car comes to a stop? Solution: Let’s first gather what we know before going further. Let p ( t ) represent the po- sition (in meters) of the car at time t (in seconds). We can fix the position at time 0 to be p (0) = 0 and in fact, suppose that t = 0 is the time when the brakes are applied. We have p 0 (0) = 50 3 (since 60 km h = 60000 m 3600 s = 50 m 3 s ), and p 00 ( t ) = - 8 for t [0 ,t stop ] . We define this special time, t stop , to be such that p 0 ( t stop ) = 0 . Let’s proceed. A particular antiderivative of - 8 is - 8 t . Obviously, p 0 is also an antiderivative. So there is a constant C such that p 0 ( t ) = - 8 t + C . Evaluating at 0 , we find that 50 3 = p 0 (0) = C , so p 0 ( t ) = 50 3 - 8 t. Now 0 = p 0 ( t stop ) = 50 3 - 8 t stop , so t stop = 50 24 = 25 12 . The distance travelled p ( t ) is an antiderivative of p 0 ( t ) . But so is 50 3 t - 4 t 2 . So they differ by a constant, that is, there exists C such that p ( t ) = 50 3 t - 4 t 2 + C . This constant is easily
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This note was uploaded on 01/18/2012 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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asst9-f10.solns - MATH 137 Assignment 9, Part 2 Fall 2010...

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