T1a1s - Assignment 1 solutions (Fall term Physics 102) 1.2...

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1.2 r, a, b, c and s all have units of L. So, () ( ) 2 LLL L L sasbsc s −−− × × == = L Assignment 1 solutions (Fall term Physics 102) 1.36 The x distance out to the fly is 2.0 m and the y distance up to the fly is 1.0 m. Thus, we can use the Pythagorean theorem to find the distance from the origin to the fly as, 22 = 2.0 m 1.0 m = dxy =+ + 2.2 m 26.6 o = = ) 0 . 2 / 0 . 1 ( tan 1 m m θ 2.12 (a) At the end of the race, the tortoise has been moving for time t and the hare for a time . The speed of the tortoise is 2.0 min 120 s tt −= 0.100 m s t v , and the speed of the hare is = 20 2.0 m s ht vv = = . The tortoise travels distance t x , which is 0.20 m larger than the distance h x traveled by the hare. Hence, , which becomes 0.20 m th xx ( ) 120 s 0.20 m vt v t = −+ or ( ) ( )( ) 0.100 m s 2.0 m s 120 s 0.20 m =− + . This gives the time of the race as t = 2 1.3 10 s × . (b) ( )( ) 2 0.100 m s 1.3 xv t × = 13 m 2.37 At the end of the acceleration period, the velocity is ( ) ( ) 2 01 . 5 m s5 . 0 s7 . 5 m s fi vva t =+=+ = . This is also the initial velocity for the braking period.

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This note was uploaded on 01/15/2011 for the course PHYS 102 taught by Professor Vandenberg during the Fall '08 term at University of Victoria.

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T1a1s - Assignment 1 solutions (Fall term Physics 102) 1.2...

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