HW02-solutions

# HW02-solutions - markowitz(am45362 – HW02 – distler –(56295 1 This print-out should have 24 questions Multiple-choice questions may continue

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Unformatted text preview: markowitz (am45362) – HW02 – distler – (56295) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points The position versus time for a certain object moving along the x-axis is shown. The ob- ject’s initial position is- 3 m.- 10- 8- 6- 4- 2 2 4 6 8 1 2 3 4 5 6 7 8 9 b b b b b b time (s) position(m) Find the instantaneous velocity at 1 s. Correct answer: 5 m / s. Explanation: The instantaneous velocity is the slope of the tangent line at that point. v = 7 m- (- 3 m) 2 s- 0 s = 5 m / s . 002 (part 2 of 4) 10.0 points Find the instantaneous velocity at 7 s. Correct answer:- 6 m / s. Explanation: v =- 9 m- (3 m) 8 s- 6 s =- 6 m / s . 003 (part 3 of 4) 10.0 points Find the average velocity between 0 s and 3 s. Correct answer: 2 m / s. Explanation: Average velocity is the change in position between the two points divided by the change in time between the two points. v = 3 m- (- 3 m) 3 s- 0 s = 2 m / s . 004 (part 4 of 4) 10.0 points Find the average velocity over the whole time shown. Correct answer: 1 m / s. Explanation: v = 6 m- (- 3 m) 9 s- 0 s = 1 m / s . 005 10.0 points A body moving with uniform acceleration has a velocity of 9 . 91 cm / s when its x coordinate is 1 . 99 cm. If its x coordinate 2 . 03 s later is- 3 . 57 cm, what is the x-component of its acceleration? Correct answer:- 12 . 462 cm / s 2 . Explanation: Let : v = 9 . 91 cm / s , x =- 3 . 57 cm , x = 1 . 99 cm , and t = 2 . 03 s . The distance is x = x + v t + 1 2 a t 2 a = 2 ( x- x- v t ) t 2 = 2 [- 3 . 57 cm- 1 . 99 cm- (9 . 91 cm / s) (2 . 03 s)] (2 . 03 s) 2 =- 12 . 462 cm / s 2 . 006 (part 1 of 2) 10.0 points A ball is thrown upward. After reaching a markowitz (am45362) – HW02 – distler – (56295) 2 maximum height, it continues falling back to- ward Earth. On the way down, the ball is caught at the same height at which it was thrown upward. b b b b b b b b b bb b b b b b b b b b v h max If the time ( up and down ) the ball re- mains in the air is 1 . 84 s, find its speed when it caught. The acceleration of grav- ity is 9 . 8 m / s 2 . Neglect air resistance. Correct answer: 9 . 016 m / s. Explanation: Let : v f =- v , Δ t = 1 . 84 s , and g =- 9 . 8 m / s 2 . v f = v + g Δ t- v = v + g Δ t Δ t =- 2 v g v =- g Δ t 2 =- (- 9 . 8 m / s 2 ) (1 . 84 s) 2 = 9 . 016 m / s . 007 (part 2 of 2) 10.0 points If the time the ball remains in the air is 1 . 97 s, find the maximum height h max the ball at- tained while in the air. Correct answer: 4 . 7541 m. Explanation: Consider the motion upward. Let : t up = 0 . 985 s and v f = 0 m / s . v f = v + g t up = 0 v =- g t up so Δ y = v t up + 1 2 g t 2 up =- g t 2 up + g t 2 up 2 =- g t 2 up 2 =- (- 9 . 8 m / s 2 ) (1 . 97 s) 2 2 = 4 . 7541 m ....
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## This note was uploaded on 09/19/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas at Austin.

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HW02-solutions - markowitz(am45362 – HW02 – distler –(56295 1 This print-out should have 24 questions Multiple-choice questions may continue

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