Homework 3−SolutionCHANG1 Take the starting pint of the patrolman as the origin. The final position for thecar and patrolman should be the same and it isxPxC2000 mBoth the car and patrolman also should have the same total traveling time, and it can beobtained from the car’s motion. Motion of the car is just single stage with starting positionx0200 m and constant velocityC36 m/s. The time of travel can be calculated to bexCx0Ct200020036tt50 sThe patrolman’s motion is multi-staged. For the first stage, we can find the velocity andposition at the transition pointAusing00 andx00:A0a1t102.5T2.5TxAx00t112a1t1200122.5T21.25T2For the second stage, the patrolman’s motion is constant velocity and his final position is at2000 m, so we havexPxAAt21.25T22.5Tt2The duration for the second stage is simplyt2t−t150−TSo the first stage’s duration can now be found to be20001.25T22.5T50−TT20 and 80Since the total time is only 50 s, the only reasonable answer isT20 sCHANG2 Take the starting point of carAas the origin, we then havex00 for the car butx0350 m for truckBatt0. Both vehicles have the same traveling time,t, and are at thesame position,xAxBwhen they meet on the road. The position of the car can be writtenusing the constant acceleration motion with zero initial velocity:xAx00t12at212at21210a25t2mThe position of the truck can also be found from the constant velocity motion:xBx0Bt350−15tmNote that the velocity of the truck is negative because it is moving in the opposite directionof the car. We then set the positions equal to find the time of travel:xAxB5t2350−15tt−10 (discard),t7 sThe distance can now be calculated to bexAxB572245 mY&F3.60 (a) The point where the snowball strikes the ground depends on the time of
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