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Unformatted text preview: yang (ey942) HW01 Tsoi (58160) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points Consider the plot below describing motion along a straight line with an initial position of 10 m. 1 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 b b b b b time (s) velocity(m/s) What is the acceleration at 1 second? Correct answer: 0 . 5 m / s 2 . Explanation: The slope of the velocity curve from 0 sec onds to 2 seconds is a = v f v i t f t i = 1 m / s 0 m / s 2 s 0 s = . 5 m / s 2 . 002 (part 2 of 3) 10.0 points What is the velocity at 2 seconds? Correct answer: 1 m / s. Explanation: The velocity at 2 seconds can be read from the plot; however, it can also be calculated: v = v i + a ( t f t i ) = 0 m / s + (0 . 5 m / s 2 ) (2 s 0 s) = 1 m / s . 003 (part 3 of 3) 10.0 points What is the position at 2 seconds? Correct answer: 11 m. Explanation: Let : x = 10 m . b b b b b b b b 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 1 time (s) velocity(m/s) The position at 2 seconds is x plus the area of the triangle (shown in gray) x = 10 m + 1 2 (2 s 0 s)(1 m / s 0 m / s) = 11 m ; however, it can also be calculated: x = x i + v i ( t f t i ) + 1 2 a ( t f t i ) 2 = 10 m + (0 m / s) (2 s 0 s) + 1 2 (0 . 5 m / s 2 ) (2 s 0 s) 2 = 11 m . 004 10.0 points A student throws a baseball at a large gong 68 m away and hears the sound of the gong 1 . 59382 s later. The speed of sound in air is 330 m / s. What was the average speed of the base ball on its way to the gong? (For simplicity, assume its trajectory to be a straight line.) Correct answer: 49 m / s. yang (ey942) HW01 Tsoi (58160) 2 Explanation: Let : d = 68 m , t = 1 . 59382 s , and v = 330 m / s . It takes t s = d v s = 68 m 330 m / s = 0 . 206061 s for the sound to travel back from the gong, so the flight time of the baseball was only t b = t tot t s = 1 . 59382 s . 206061 s = 1 . 38776 s and the baseballs average speed was v b = d t = 68 m 1 . 38776 s = 49 m / s . 005 (part 1 of 3) 10.0 points A boy is standing on a cliff. The boys hands are a height b above the ground level at the base of the cliff. A monkey is in a tree. The monkey is at a height h above the boys hands. At t = 0 the boy throws a coconut upward at a speed v , and at the same time the monkey releases his grip, falling downward to catch the coconut. Assume the initial speed of the monkey is 0 m/s, and the cliff is high enough so that the monkey is able to catch the coconut before hitting the ground. The acceleration of gravity is 10 m / s 2 . b h A ground level B hand height C monkey height v Figure: Consider only vertical motion....
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 Fall '10
 TSIO

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