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HW 2-solutions - mitra(am53956 HW 2 Opyrchal(11113 This...

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mitra (am53956) – HW 2 – Opyrchal – (11113) 1 This print-out should have 17 questions. Multiple-choice 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 11 mi at 52 mi / h, then 25 . 4 mi at 43 mi / h, and finally 48 . 5 mi at 38 mi / h. What is the car’s average velocity during the entire trip? Correct answer: 40 . 8457 mi / h. Explanation: Let : d A = 11 mi , v A = 52 mi / h , d B = 25 . 4 mi , v B = 43 mi / h , d C = 48 . 5 mi , and v C = 38 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 = 11 mi 52 mi / h + 25 . 4 mi 43 mi / h + 48 . 5 mi 38 mi / h = 2 . 07855 h , so the average velocity is v = Δ d Δ t = d A + d B + d C Δ t = 11 mi + 25 . 4 mi + 48 . 5 mi 2 . 07855 h = 40 . 8457 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 0 = 10 m. - 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 time (s) velocity (m/s) What is the position at 2 seconds? Correct answer: 19 m. Explanation: The initial position given in the problem is 10 m.
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mitra (am53956) – HW 2 – Opyrchal – (11113) 2 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 - 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) × (9 m / s - 0 m / s) = 19 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 (4 . 5 m / s 2 ) (2 s - 0 s) 2 = 19 m . 003 (part 2 of 3) 10.0 points What is the position at 6 seconds? Correct answer: 43 m. Explanation: The position is 19 m plus the area of the trapezoid from 2 s to 6 s x = 19 m + 1 2 (6 s - 2 s) × (3 m / s + 9 m / s) = 43 m ; however, it can also be calculated: x = x i + v i ( t f
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