25_Physics ProblemsTechnical Physics

25_Physics ProblemsTechnical Physics - 26 Motion in One...

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26 Motion in One Dimension Section 2.3 Acceleration P2.11 Choose the positive direction to be the outward direction, perpendicular to the wall. vv a t fi =+ : a v t == −− × 22 0 25 0 350 10 134 10 3 4 .. . . ms s ±ms 2 af . P2.12 (a) Acceleration is constant over the first ten seconds, so at the end, a t =+=+ ( ) = 0 2 00 10 0 20 0 . s 2 ch . Then a = 0 so v is constant from t = 10 0 . s to t = 15 0 . s. And over the last five seconds the velocity changes to a t =+= + ( ) = 20 0 3 00 5 00 5 00 . . s 2 . (b) In the first ten seconds, xx v t a t fii =+ + =++ ( )= 1 2 00 1 2 200 100 2 2 s m 2 . Over the next five seconds the position changes to v t a t = + ( ) += 1 2 500 0 200 2 m m s s m . And at t = 20 0 . s , v t a t = + ( ) + () = 1 2 1 2 3 00 5 00 262 2 2 m m s s m s s m 2 . . afch . *P2.13 (a) The average speed during a time interval t is v t = distance traveled
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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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