PHSX 114 - HW #2 Solutions

PHSX 114 - HW #2 Solutions - = +-= + = + =-( b )The time of...

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Homework Assignment 3 - Solutions 21. By definition, the acceleration is 2 0 25m s 13m s 2.0 m s 6.0 s v v a t - - = = = . The distance of travel can be found from Eq. 2-11b. ( 29 ( 29 ( 29 ( 29 2 2 2 1 1 0 0 2 2 13m s 6.0 s 2.0 m s 6.0 s 114 m x x v t at - = + = + = 26. The final velocity of the car is zero. The initial velocity is found from Eq. 2-11c with 0 v = and solving for 0 v . ( 29 ( 29 ( 29 ( 29 2 2 2 2 0 0 0 0 2 2 0 2 7.00 m s 92 m 36 m s v v a x x v v a x x = + - = - - = - - = 36. Choose upward to be the positive direction, and take 0 0 y = to be at the height where the ball was hit. For the upward path, 0 22 m s v = , 0 v = at the top of the path, and 2 9.80 m s a = - . ( a ) The displacement can be found from Eq. 2-11c, with x replaced by y . ( 29 ( 29 ( 29 2 2 2 2 2 0 0 0 0 2 0 22m s 2 0 25 m 2 2 9.80m s v v v v a y y y y a - -
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Unformatted text preview: = +-= + = + =-( b )The time of flight can be found from Eq. 2-11b, with x replaced by y , using a displacement of 0 for the displacement of the ball returning to the height from which it was hit. ( 29 ( 29 2 1 1 2 2 2 2 22m s 2 0 0 0 , 4.5 s 9.80m s v y y v t at t v at t t a = + + = + = = = = =--The result of t = 0 s is the time for the original displacement of zero (when the ball was hit), and the result of t = 4.5 s is the time to return to the original displacement. Thus the answer is t = 4.5 seconds....
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