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Exam1-Solutions-S14

# That rules out 2 leaving just 1 problem 8 one

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Unformatted text preview: ng in the positive x direction at t =0. For that car the kinematic equation becomes: x P = 0 + Pt . At that instant the second car crosses a point at x = d km going in the –x direction with velocity –Q. For that car x Q = d − Qt . To find the time when they meet we equate xP and xQ: yielding for the time when the two cars meet: t = d / ( P + Q) . Here P + Q = 100km / h so 200, 250 or 300 km t= = 2, 2.5 or 3 h 100km / h PHY 2048 Spring 2014 – Acosta, Rinzler Exam 1 solutions Problem 7 A car accelerates from rest on a straight road. A short time later, the car decelerates to a stop and then returns to its original position in a similar manner, by speeding up and then slowing to a stop. Which of the following five coordinate versus time graphs best describes the motion? The car goes out and comes back to th...
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