Solution TAM 212 Sample Exam number 1, Chapter 1; Monday Sept 13, 2004. 1] A particle increases its speed along its path at a uniform rate of 1m/sec 2 from its original position at rest at point A. Its path is straight for 8 meters and parabolic thereafter. Find the speed at B. Also find the magnitude of its acceleration vector | a | immediately after entering the curved section at B. Note that its speed is still increasing after the particle passes B. Answer: 8 meters = (1/2) a t 2 ; s=with a = 1 so t = 4 seconds and its speed then is 4 m/sec when it gets to point B. [ alternatively you could write the familiar(?) v = sqrt(2a d) ] Just after it passes B it is moving horizontally with (now we use path variables) s-dot = 4m/sec, s-doubledot = 1m/sec 2 (recall it is still increasing its speed) and rho = radius of curvature = (1+y’ 2 ) 3/2 /|y”| = 10 meters. So a = 1 e t + (4) 2 /10 e n = 1 e t + 1.6 e n . And | a | = sqrt(1+1.6 2 ) = 1.89 m/sec 2 . 2]
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