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Unformatted text preview: Patel (pnp223) homework 13 Turner (92160) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A hunter wishes to cross a river that is 1 . 7 km wide and that flows with a speed of 6 . 2 km / h. The hunter uses a small powerboat that moves at a maximum speed of 12 km / h with respect to the water. What is the time necessary for crossing if the boat moves directly across the river? Draw the vectors to scale on a graph to deter mine the answer. Correct answer: 9 . 92774 min. Explanation: Let : x = 1 . 7 km , v re = 6 . 2 km / h , and v br = 12 km / h . v be 6 . 2 km / h 1 2 k m / h Scale: 10 km / hr We need to find the angle to head into the current so that the resultant direction is perpendicular to the current v br sin = v re sin = v re v br = sin 1 parenleftbigg v re v br parenrightbigg = sin 1 parenleftbigg 6 . 2 km / h 12 km / h parenrightbigg = 31 . 1089 . Thus v be = v br cos = (12 km / h) cos(31 . 1089 ) = 10 . 2742 km / h . and t min = x v be = x v br cos = 1 . 7 km (12 km / h) cos(31 . 1089 ) 60 min 1 h = 9 . 92774 min . Alternate Solution: v be = radicalBig v 2 be v 2 re t min = x v be = (1 . 7 km) radicalbig (10 . 2742 km / h) 2 (6 . 2 km / h) 2 = (1 . 7 km) (10 . 2742 km / h) 60 min 1 h = 9 . 92774 min . 002 10.0 points A car is moving at constant speed on the freeway. The driver sees a patrol car at time t 1 and rapidly slows down by around 10 miles per hour. After continuing at this speed for a few minutes, the driver at time t 2 returns to the earlier constant speed. Let us plot the acceleration of the car as a function of time; take the forward direction of motion as positive. Which graph correctly describes the cars acceleration a ( t )? 1. t 1 time a t 2 Patel (pnp223) homework 13 Turner (92160) 2 2. t 1 time a t 2 3. t 1 time a t 2 correct 4. t 1 time a t 2 5. t 1 time a t 2 6. t 1 time a t 2 7. t 1 time a t 2 8. t 1 time a t 2 Explanation: The car is at first moving at a constant speed and therefore its acceleration is zero....
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This note was uploaded on 07/11/2009 for the course PHY 92160 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Turner
 Physics, Work

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