midterm review - brown(twb493 – midterm 01 rev – turner...

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Unformatted text preview: brown (twb493) – midterm 01 rev – turner – (57340) 1 This print-out should have 33 questions. Multiple-choice 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.

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midterm review - brown(twb493 – midterm 01 rev – turner...

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