Solutions02 - Physics 140 - Discussion 02 - Solutions 2.23....

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Physics 140 - Discussion 02 - Solutions 2.23. I DENTIFY : Assume that the acceleration is constant and apply the constant acceleration kinematic equations. Set | | x a equal to its maximum allowed value. S ET U P : Let x + be the direction of the initial velocity of the car. a x = - 250 m/s 2 . v 0x = + 105 km/h 29 17 m/s. = . E XECUTE : v 0x = + 29.17 m/s, 0. x v = 2 2 0 0 2 ( ) x x x v v a x x = + - gives 2 2 2 0 0 2 0 (29 17 m/s) 1 70 m. 2 2( 250 m/s ) x x x v v x x a - - . - = = = . - E VALUATE : The car frame stops over a shorter distance and has a larger magnitude of acceleration. Part of your 1.70 m stopping distance is the stopping distance of the car and part is how far you move relative to the car while stopping. ______________________________________________________________________________ 2.42. I DENTIFY : Apply constant acceleration equations to the vertical motion of the brick. S ET U P : Let y + be downward. 2 9 80 m/s y a = . E XECUTE : (a) 2 0 0, 2 50 s, 9 80 m/s . y y v t a = = . = . 2 2 2 1 1 0 0 2 2 (9 80 m/s )(2 50 s) 30 6 m. y y y y v t a t - = + = . . = . The building is 30.6 m tall. (b) 2 0 0 (9 80 m/s )(2 50 s) 24 5 m/s y y y v v a t = + = + . . = . (c) The graphs of , y a y v and y versus t are given in Figure 2.42. Take 0 y = at the ground. E
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This note was uploaded on 02/15/2012 for the course PHYS 140 taught by Professor Uher during the Winter '10 term at University of Michigan.

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Solutions02 - Physics 140 - Discussion 02 - Solutions 2.23....

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