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Unformatted text preview: kuruvila (lk5992) – HW 2 – Opyrchal – (11113) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A car travels along a straight stretch of road. It proceeds for 13 . 6 mi at 58 mi / h, then 23 . 7 mi at 46 mi / h, and finally 46 . 5 mi at 36 . 9 mi / h. What is the car’s average velocity during the entire trip? Correct answer: 41 . 6944 mi / h. Explanation: Let : d A = 13 . 6 mi , v A = 58 mi / h , d B = 23 . 7 mi , v B = 46 mi / h , d C = 46 . 5 mi , and v C = 36 . 9 mi / h . The total time the car spent on the road is Δ t = d A v A + d B v B + d C v C = 13 . 6 mi 58 mi / h + 23 . 7 mi 46 mi / h + 46 . 5 mi 36 . 9 mi / h = 2 . 00986 h , so the average velocity is v = Δ d Δ t = d A + d B + d C Δ t = 13 . 6 mi + 23 . 7 mi + 46 . 5 mi 2 . 00986 h = 41 . 6944 mi / h . 002 (part 1 of 3) 10.0 points Consider the plot below describing motion along a straight line with an initial position of x = 10 m. 4 3 2 1 1 2 3 1 2 3 4 5 6 7 8 9 b b b b b time (s) velocity(m/s) What is the position at 2 seconds? Correct answer: 11 m. Explanation: The initial position given in the problem is 10 m. b b b b b b b b 1 2 3 4 5 6 7 8 9 1 2 3 1 2 3 4 time (s) velocity(m/s) The position at 2 seconds is 10 meters plus the area of the triangle (shaded in the above plot) 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 ( 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 . kuruvila (lk5992) – HW 2 – Opyrchal – (11113) 2 003 (part 2 of 3) 10.0 points What is the position at 6 seconds?...
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This note was uploaded on 10/22/2010 for the course PHYS 111 taught by Professor Moro during the Fall '08 term at NJIT.
 Fall '08
 moro

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