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Physics hw 2 solutions

# Physics hw 2 solutions - mata(jpm2873 HW#2...

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mata (jpm2873) – HW #2 – Antoniewicz – (57420) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of5)2.0points Consider a toy car which can move to the right (positive direction) or left on a horizon- tal surface along a straight line. car v O + What is the acceleration-time graph if the car moves toward the right (away from the origin), speeding up at a steady rate? 1. t a 2. t a 3. t a 4. None of these graphs is correct. 5. t a 6. t a 7. t a correct 8. t a Explanation: Since the car speeds up at a steady rate, the acceleration is a constant. 002(part2of5)2.0points What is the acceleration-time graph if the car moves toward the right, slowing down at a steady rate? 1. t a 2. t a 3. t a

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mata (jpm2873) – HW #2 – Antoniewicz – (57420) 2 4. t a 5. t a correct 6. t a 7. None of these graphs is correct. 8. t a Explanation: Since the car slows down, the acceleration is in the opposite direction. 003(part3of5)2.0points What is the acceleration-time graph if the car moves towards the left (toward the origin) at a constant velocity? 1. t a 2. t a 3. t a 4. None of these graphs is correct. 5. t a 6. t a correct 7. t a 8. t a Explanation: Since the car moves at a constant velocity, the acceleration is zero. 004(part4of5)2.0points What is the acceleration-time graph if the car moves toward the left, speeding up at a steady rate?
mata (jpm2873) – HW #2 – Antoniewicz – (57420) 3 1. t a correct 2. t a 3. t a 4. t a 5. t a 6. t a 7. None of these graphs is correct. 8. t a Explanation: The same reason as Part 1. 005(part5of5)2.0points What is the acceleration-time graph if the car moves toward the right at a constant velocity? 1. t a 2. t a 3. t a correct 4. None of these graphs is correct. 5. t a 6. t a 7. t a

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mata (jpm2873) – HW #2 – Antoniewicz – (57420) 4 8. t a Explanation: The same reason as Part 3. 006(part1of3)4.0points A train travels between stations 1 and 2, as shown in the figure. The engineer of the train is instructed to start from rest at station 1 and accelerate uniformly between points A and B, then coast with a uniform velocity be- tween points B and C, and finally decelerate uniformly between points C and D until the train stops at station 2. The distances AB, BC, and CD are all equal, and it takes 4 . 02 min to travel between the two stations. As- sume that the uniform accelerations have the same magnitude, even when they are opposite in direction. Station A Station B A B C D How much of this 4 . 02 min period does the train spend between points A and B? Correct answer: 1 . 608 min. Explanation: Δ x AB = Δ x BC = Δ x CD Δ t AB + Δ t BC + Δ t CD = 4 . 02 min v A = v D = 0 m / s v = Δ x Δ t = v o + v f 2 Δ t = Δ x v = 2 Δ x v o + v f . Between points A and B, Δ t AB = 2 Δ x AB v B . Because the train’s velocity is constant from B to C, Δ t BC = Δ x BC v B = Δ x AB v B .
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