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Unformatted text preview: brown (twb493) – midterm 01 rev – turner – (57340) 1 This printout should have 33 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Two cars are traveling along a straight line in the same direction, the lead car at 33 m / s and the other car at 48 m / s. At the moment the cars are 47 m apart, the lead driver applies the brakes, causing the car to have a deceleration of 1 . 5 m/s 2 . How long does it take for the lead car to stop? Correct answer: 22 s. Explanation: Let : v i, 1 = +33 m / s , v f, 1 = 0 m / s , and a 1 = − 1 . 5 m / s 2 . a 1 = v f, 1 − v i, 1 Δ t 1 = − v i, 1 Δ t 1 Δ t 1 = − v i, 1 a 1 = − 33 m / s − 1 . 5 m / s 2 = 22 s . 002 (part 2 of 3) 10.0 points Assume that the driver of the chasing car applies the brakes at the same time as the driver of the lead car. What must the chasing car’s minimum neg ative acceleration be to avoid hitting the lead car? Correct answer: − 2 . 80976 m / s 2 . Explanation: Let : v i, 2 = +48 m / s , v f, 2 = 0 m / s , Δ x = 47 m , and Δ t 2 = Δ t 1 = 22 s . The lead car traveled Δ x 1 = 1 2 ( v f, 1 + v i, 1 ) Δ t 1 = v i, 1 2 Δ t 1 = 33 m / s 2 (22 s) = 363 m . The chasing car traveled Δ x 2 = Δ x 1 + Δ x = 363 m + 47 m = 410 m with an acceleration defined by 0 = v 2 i, 2 + 2 a 2 Δ x 2 a 2 = − v 2 i, 2 2 Δ x 2 = − (48 m / s) 2 2(410 m) = − 2 . 80976 m / s 2 . 003 (part 3 of 3) 10.0 points How long does it take the chasing car to stop? Correct answer: 17 . 0833 s. Explanation: a 2 = v f, 2 − v i, 2 Δ t 2 = − v i, 2 Δ t 2 Δ t 2 = − v i, 2 a 2 = − 48 m / s − 2 . 80976 m / s 2 = 17 . 0833 s . 004 (part 1 of 2) 10.0 points Throw a ball upward from point O with an initial speed of 82 m / s. t h A y t B h B t A b b b b b b b b b b b b b b b b b b b b b b b b 82 m / s O B A What is the maximum height? The accel eration of gravity is 9 . 8 m / s 2 . brown (twb493) – midterm 01 rev – turner – (57340) 2 Correct answer: 343 . 061 m. Explanation: Let up be positive. Let : v = 82 m / s , v f = 0 , and g = − 9 . 8 m / s 2 . v 2 f = v 2 + 2 g Δ h = 0 h A = − v 2 2 g = (82 m / s) 2 2 ( − 9 . 8 m / s 2 ) = 343 . 061 m . 005 (part 2 of 2) 10.0 points If the speed of the ball as it passes point B is 1 2 (82 m / s) = 41 m / s, what is the height h B of B above O ? Correct answer: 257 . 296 m. Explanation: parenleftBig v 2 parenrightBig 2 − v 2 = − 2 g h B − 3 4 v 2 o = − 2 g h B 3 4 ( − 2 g h A ) = 2 g h B h B = 3 4 h A = 3 4 (343 . 061 m) = 257 . 296 m . 006 10.0 points Michael stands motionless holding a base ball in his hand. After a while he tosses it upwards, and it travels up for a while be fore turning about and heading toward the ground. Define upwards to be positive....
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This note was uploaded on 04/13/2011 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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