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HW 02 Solution

HW 02 Solution - Question 1 chap 2 sect 2 part 1 of 1 10...

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Question 1, chap 2, sect 2. part 1 of 1 10 points The graph below shows the velocity v as a function of time t for an object moving in a straight line. t v 0 t Q t R t S t P Which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval? 1. t x 0 t Q t R t S t P 2. t x 0 t Q t R t S t P 3. t x 0 t Q t R t S t P 4. t x 0 t Q t R t S t P 5. t x 0 t Q t R t S t P 6. t x 0 t Q t R t S t P correct 7. None of these graphs are correct. 8. t x 0 t Q t R t S t P 9. t x 0 t Q t R t S t P Explanation: The displacement is the integral of the ve- locity with respect to time: vectorx = integraldisplay vectorv dt . Because the velocity increases linearly from zero at first, then remains constant, then de- creases linearly to zero, the displacement will increase at first proportional to time squared, then increase linearly, and then increase pro- portional to negative time squared. From these facts, we can obtain the correct answer. t x 0 t Q t R t S t P Question 2, chap 2, sect 2. part 1 of 2 10 points You drive a car 9 h at 46 km / h, then 9 h at 66 km / h. What is your average velocity? Correct answer: 56 km / h (tolerance ± 1 %). Explanation: Basic Concept Average velocity is defined by v av = Δ d Δ t Solution If you drive for t hours at a speed of v 1 and for the same time at a speed of v 2 , then Δ d = d 1 + d 2 = v 1 t + v 2 t in a time of Δ t = t 1 + t 2 = 2 t

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so that the average velocity is v av = v 1 t + v 2 t 2 t = v 1 + v 2 2 = (46 km / h) + (66 km / h) 2 = 56 km / h . Question 3, chap 2, sect 2. part 2 of 2 10 points What is your average velocity if you drive a distance of 504 km at a speed of 46 km / h, then the same distance at a speed of 66 km / h? Correct answer: 54 . 2143 km / h (tolerance ± 1 %). Explanation: If you drive the distance d at v 1 and the same distance at v 2 , then Δ d = d 1 + d 2 = 2 d and Δ t = t 1 + t 2 = d v 1 + d v 2 so that the average velocity will be v av = 2 d d v 1 + d v 2 · v 1 v 2 v 1 v 2 = 2 d v 1 v 2 d v 2 + d v 1 = 2 v 1 v 2 v 1 + v 2 = 2 (46 km / h) (66 km / h) (46 km / h) + (66 km / h) = 54 . 2143 km / h . Question 4, chap 2, sect 2. part 1 of 2 10 points A particle moves according to the equation x = (10 m / s 2 ) t 2 where x is in meters and t is in seconds. Find the average velocity for the time in- terval from t 1 = 2 . 16 s to t 2 = 4 . 63 s. Correct answer: 67 . 9 m / s (tolerance ± 1 %). Explanation: x = (10 m / s 2 ) t 2 v ave = Δ x Δ t = (10 m / s 2 ) ( t 2 2 - t 2 1 ) t 2 - t 1 = (10 m / s 2 ) [(4 . 63 s) 2 - (2 . 16 s) 2 ] (4 . 63 s) - (2 . 16 s) = 67 . 9 m / s Question 5, chap 2, sect 2. part 2 of 2 10 points Find the average velocity for the time in- terval from t 1 = 2 . 16 s to t 3 = 2 . 26 s. Correct answer: 44 . 2 m / s (tolerance ± 1 %). Explanation: v ave = Δ x Δ t = (10 m / s 2 ) ( t 2 3 - t 2 1 ) t 3 - t 1 = (10 m / s 2 ) [(2 . 26 s) 2 - (2 . 16 s) 2 ] (2 . 26 s) - (2 . 16 s) = 44 . 2 m / s Question 6, chap 2, sect 2. part 1 of 1 10 points A car travels along a straight stretch of road. It proceeds for 12 . 5 mi at 51 mi / h, then 25 . 1 mi at 45 mi / h, and finally 39 . 6 mi at 35 . 4 mi / h.
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HW 02 Solution - Question 1 chap 2 sect 2 part 1 of 1 10...

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