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Unformatted text preview: Version 046/AACDC – midterm 01 – Turner – (58220) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A boat moves through the water of a river at 14 . 9 m / s relative to the water, regardless of the boat’s direction. If the current is flowing at 11 . 9 m / s, how long does it take the boat to complete a trip consisting of a 432 m displacement down stream followed by a 171 m displacement up stream? 1. 359.125 2. 193.712 3. 292.592 4. 197.317 5. 258.751 6. 86.8937 7. 73.1194 8. 72.14 9. 147.094 10. 183.559 Correct answer: 73 . 1194 s. Explanation: Let v b be the velocity of the boat relative to the water, and v w the velocity of water relative to the shore. Take downstream as the positive direction. For the downstream trip, the current is speeding him up, so the velocity of the boat relative to the shore is v d = v b + v w and the time is t d = x d v d = x d v b + v w For the upstream trip, the current is slowing him down, so the velocity of the boat relative to the shore is v u = v b − v w and the time is t u = x u v u = x u v b − v w The time for the entire trip is thus t = t d + t u = x d v b + v w + x u v b − v w = (432 m) (14 . 9 m / s) + (11 . 9 m / s) + (171 m) (14 . 9 m / s) − (11 . 9 m / s) = 73 . 1194 s 002 10.0 points Jimmy is at the bottom of a hill while Billy is on the hill a distance 17 m up the hill from the bottom. Jimmy is at the origin of an xy coordinate system, and the line that follows the slope of the hill has an equation y = (0 . 53) x . Jimmy throws an apple to Billy at an angle of 48 ◦ with respect to the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 48 ◦ y = . 5 3 x v 1 7 m Billy Jimmy y x With what speed v must he throw the apple to reach Billy? 1. 20.8571 2. 17.0418 3. 19.0808 4. 21.3574 5. 23.5333 6. 24.9482 7. 16.8263 8. 20.1547 9. 22.2633 10. 16.1939 Correct answer: 16 . 8263 m / s. Explanation: Let : θ = 48 ◦ , Version 046/AACDC – midterm 01 – Turner – (58220) 2 d = 17 m , and b = 0 . 53 , where y = b x . The kinematic equations in the x and y directions give x = ( v cos θ ) t (1) y = ( v sin θ ) t − 1 2 g t 2 . (2) Eliminating t in Eq. 2 with t = x v cos θ yields y = v x sin θ v cos θ − g x 2 2 v 2 cos 2 θ = x tan θ − g x 2 2 v 2 cos 2 θ . Solving for v we obtain v = radicalBigg g x 2 2 cos 2 θ ( x tan θ − y ) . (3) Billy’s location is at ( x, y ) and the distance separating Billy and Jimmy is given by d = 17 m , so d = radicalbig x 2 + y 2 = radicalbig x 2 + b 2 x 2 = x radicalbig 1 + b 2 . Solving for x and y = b x , and plugging in the appropriate values yields x = d √ 1 + b 2 (4) = (17 m) radicalbig 1 + (0 . 53) 2 = 15 . 0207 m , and y = b x = b d √ 1 + b 2 (5) = (0 . 53) (17 m) radicalbig 1 + (0 . 53) 2 = 7 . 96099 m ....
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This note was uploaded on 04/07/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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