23_Physics ProblemsTechnical Physics

# 23_Physics - 24 P2.5 Motion in One Dimension(a Let d represent the distance between A and B Let t1 be the time for which the walker has d the

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24 Motion in One Dimension P2.5 (a) Let d represent the distance between A and B. Let t 1 be the time for which the walker has the higher speed in 500 1 . m s = d t . Let t 2 represent the longer time for the return trip in −= 300 2 m s d t . Then the times are t d 1 = . m s bg and t d 2 = m s . The average speed is: v dd d v d == + + = Total distance Total time ms 22 800 15 0 2 2150 375 .. . . . . . ej (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0. Section 2.2 Instantaneous Velocity and Speed P2.6 (a) At any time, t , the position is given by xt = 2 m s 2 . Thus, at t i = s : x i 3 00 3 00 27 0 2 . s m 2 a f . (b) At tt f =+ s : f 2 ±ms ± s 2 af , or t f + 270 180 2 . m m s m s 2 ∆∆ . (c) The instantaneous velocity at t = s is: v xx t t t fi t = F H G I K J = →→ lim lim . . . 00 18 0 3 00 18 0 2 .
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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