02_P47InstructorSolution

02_P47InstructorSolution - 2.47. Model: Visualize: Solve:...

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2.47. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: Solve: (a) To find 2 , x we first need to determine 1 . x Using 100 1 0 () , x xv tt =+ − we get 1 0 m (20 m/s)(0.50 s 0 s) x =+ −= 10 m. Now, 22 2 2 2 2 21 1 2 2 2 ( ) 0 m /s (20 m/s) 2( 10 m/s )( 10 m) 30 m vv a xx x x − ⇒ = +− ⇒= The distance between you and the deer is 32 x x or (35 m 30 m) 5 m. = (b) Let us find 0 max v such that 2 0 m/s v = at 23 35 m. == Using the following equation, 2 2 20 m a x 1 0 m a x 1 2 ( ) 0 m /s 2( 10 m/s )(35 m ) vv a v x Also, 1 0 0 max 1 0 0 max 0 max ( ) (0.50 s 0 s) (0.50 s) . xxv tt v v − = Substituting this expression for x 1 in the above
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at UGA.

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