a2_001 - After time t , x A =-6 . 00 + 4 . 00 t , and x B =...

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PHY 2053, Fall 2009, Quiz 2 — Whiting 1. A person takes a trip, driving with a constant speed of 90.0 km/hr, except for a 15.0-min rest stop. If the person’s average speed is 60.0 km/hr, a) How much time is spent on the trip? Let t be the total time for the trip (in hr), and let x be the total distamce travelled (in km). Then: x = 60 . 0 × t = 90 . 0 × ( t - 0 . 250) = t = 90 . 0 × 0 . 250 30 . 0 = 0 . 750 hr = 45 . 0 min . b) How far does the person travel? t = x 60 . 0 = x 90 . 0 + 0 . 250 = x = 0 . 250 × 60 . 0 × 90 . 0 (90 . 0 - 60 . 0) = 45 . 0 km 2. Runner A is initially 6.00 km west of a ±agpole and is running with a constant velocity of 4.00 km/hr due east. Runner B is initially 5.00 km east of the ±agpole and is running with a constant velocity of 3.00 km/hr due west. a) How long will it take before the runners meet?
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Unformatted text preview: After time t , x A =-6 . 00 + 4 . 00 t , and x B = 5 . 00-3 . 00 t . When they meet (at t meet ), x A = x B , i.e., x A =-6 . 00 + 4 . 00 t meet = 5 . 00-3 . 00 t meet = x B . = t meet = 11 . 00 / 7 . 00 hr = 1 . 57 hr . b) How far are they from the agpole when they meet? Consider the time it takes each to reach the position x . t A = x-(-6 . 00) 4 . 00 , t B = x-5 . 00 (-3 . 00) . When they meet (at x meet ), t A = t B . Thus: t A = x meet-(-6 . 00) 4 . 00 = x meet-5 . 00 (-3 . 00) = t B = ( 1 3 . 00 + 1 4 . 00 ) x meet = 5 . 00 3 . 00-6 . 00 4 . 00 , i . e ., x meet = 2 . 00 7 . 00 km = 0 . 286 km ....
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This note was uploaded on 12/15/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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