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# HS2 - Adams Russell Homework 2 Due midnight Inst Opyrchal H...

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Adams, Russell – Homework 2 – Due: Sep 22 2006, midnight – Inst: Opyrchal, H 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A car travels along a straight stretch of road. It proceeds for 17 . 8 mi at 58 mi / h, then 26 mi at 43 mi / h, and finally 49 . 4 mi at 37 . 4 mi / h. What is the car’s average velocity during the entire trip? Correct answer: 41 . 7487 mi / h. Explanation: The total distance the car traveled is Δ d = d A + d B + d C = 93 . 2 mi The total time the car spent on the road is Δ t = d A v A + d B v B + d C v C = 17 . 8 mi 58 mi / h + 26 mi 43 mi / h + 49 . 4 mi 37 . 4 mi / h = 2 . 2324 h . Hence the average velocity is v = Δ d Δ t = 93 . 2 mi 2 . 2324 h = 41 . 7487 mi / h . keywords: 002 (part 1 of 1) 10 points A car, moving along a straight stretch of high- way, begins to accelerate at 0 . 0271 m / s 2 . It takes the car 49 . 4 s to cover 1 km. How fast was the car going when it first began to accelerate? Correct answer: 19 . 5735 m / s. Explanation: We can describe this situation with the equation d = v 0 t + 1 2 at 2 Given the time, distance and acceleration we simply solve for the initial velocity v 0 = d t - 1 2 at converting the distance from km to m to ob- tain the proper units. keywords: 003 (part 1 of 8) 10 points Consider the plot below describing motion along a straight line with an initial position of x 0 = 10 m. - 2 - 1 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 8 9 time (s) velocity (m/s) What is the acceleration at 1 second? Correct answer: 3 . 5 m / s 2 . Explanation: Basic Concepts: The plot shows a curve of the velocity versus time . The acceleration is the slope of the velocity curve a = d v dt = Δ v Δ t .

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HS2 - Adams Russell Homework 2 Due midnight Inst Opyrchal H...

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