ch05-p011 - 11. To solve the problem, we note that...

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2 2 7.00 18.0 , 18.0 dy d y t dt dt =− = (a) The acceleration is 22 ˆˆ ˆ ˆ ˆ ˆ i j i j ( 24.0 )i ( 18.0)j. xy dx dy aa a t dt dt =+= + = + G At 0.700 s t = , we have ( 16.8)i ( 18.0)j a +− G with a magnitude of 2 | | ( 16.8) ( 18.0) 24.6 m/s . aa == + = G Thus, the magnitude of the force is 2 (0.34 kg)(24.6 m/s ) 8.37 N. Fm a = (b) The angle F G or / aFm = G G makes with x + is 2 11 2 18.0 m/s tan tan 47.0 or 133 . 16.8 m/s y x a a θ −− §· = ° ° ¨¸ ©¹ We choose the latter ( 133 −° ) since F G is in the third quadrant. (c) The direction of travel is the direction of a tangent to the path, which is the direction of the velocity vector: 2 ˆ ˆ ˆ ˆ ( ) i j i j (2.00 12.0 )i (7.00 18.0 )j. dx dy vt v v t t dt dt + = + G At 0.700 s t = , we have ( 0.700 s) ( 3.88 m/s)i ( 5.60 m/s)j. + G Therefore, the angle v G makes with x + is 5.60 m/s tan tan 55.3 or 125 . 3.88 m/s y v x v v = °
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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