Solutions03 - Physics 140 - Discussion 03 - Solutions 3.3....

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Unformatted text preview: Physics 140 - Discussion 03 - Solutions 3.3. (a) I DENTIFY and S ET U P : From r r we can calculate x and y for any t . Then use Eq. (3.2), in component form. E XECUTE : 2 2 [4.0 cm (2.5 cm/s ) ] (5.0 cm/s) = + + r t t r i j At 0, t = (4 0 cm) = . . r i r At 2 0 s, t = . (14 0 cm) (10 0 cm) . = . + . r i j r av 10.0 cm ( ) 5.0 cm/s. 2.0 s = = = x x v t av 10.0 cm ( ) 5.0 cm/s. 2.0 s = = = y y v t 2 2 av av av ( ) ( ) 7 1 cm/s x y v v v = + = . av av ( ) tan 1 00 ( ) y x v v = = . 45 . = Figure 3.3a E VALUATE : Both x and y increase, so av v r is in the 1st quadrant. (b) I DENTIFY and S ET U P : Calculate r r by taking the time derivative of ( ). t r r E XECUTE : 2 ([5.0 cm/s ] ) (5.0 cm/s) = = + r r d t dt r v i j 0: t = 0, x v = 5 0 cm/s; y v = . 5 0 cm/s v = . and 90 = 1 0 s: t = . 5 0 cm/s, x v = . 5 0 cm/s; y v = . 7 1 cm/s v = . and 45 = 2 0 s: t = . 10 0 cm/s, x v = . 5 0 cm/s; y v = . 11 cm/s v = and 27 = (c) The trajectory is a graph of y versus x . 2 2 4 0 cm (2 5 cm/s ) , x t = . + . (5.0 cm/s) = y t For values of t between 0 and 2.0 s, calculate x and y and plot y versus x . Figure 3.3b E VALUATE : The sketch shows that the instantaneous velocity at any t is tangent to the trajectory....
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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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Solutions03 - Physics 140 - Discussion 03 - Solutions 3.3....

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