# 20 - fryer(vdf96 – Kinematics – graves –(6 1 This...

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Unformatted text preview: fryer (vdf96) – Kinematics – graves – (6) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The graph shows the velocity v as a function of time t for an object moving in a straight line. t v t Q t R t S t P Which graph shows the corresponding dis- placement x as a function of time t for the same time interval? 1. t x t Q t R t S t P 2. t x t Q t R t S t P correct 3. t x t Q t R t S t P 4. t x t Q t R t S t P 5. t x t Q t R t S t P 6. None of these graphs is correct. 7. t x t Q t R t S t P 8. t x t Q t R t S t P 9. t x t Q t R t S t P Explanation: The displacement is the integral of the ve- locity with respect to time: vectorx = integraldisplay vectorv dt . Because the velocity increases linearly from zero at first, then remains constant, then de- creases linearly to zero, the displacement will increase at first proportional to time squared, then increase linearly, and then increase pro- portional to negative time squared. From these facts, we can obtain the correct answer. t x t Q t R t S t P 002 10.0 points Two ants race across a table 58 cm long. One travels at 4 . 96 cm / s and the other at 2 cm / s. When the first one crosses the finish line, how far behind is the second one? Correct answer: 34 . 6129 cm. Explanation: Let : ℓ = 58 cm , v 1 = 4 . 96 cm / s , and v 2 = 2 cm / s . The time it takes the first (faster) ant to fryer (vdf96) – Kinematics – graves – (6) 2 cross the finish line is t = ℓ v 1 , and the distance the slower ant covers in that time is s 2 = v 2 t = v 2 ℓ v 1 . The slower ant is ℓ- s 2 = ℓ- v 2 ℓ v 1 = 58 cm- (2 cm / s) (58 cm) 4 . 96 cm / s = 34 . 6129 cm from the finish line when the faster one crosses it. 003 10.0 points A runner is jogging at a steady 1 km / hr. When the runner is 0 . 88 km from the finish line a bird begins flying from the runner to the finish line at 3 km / hr (3 times as fast as the runner). When the bird reaches the finish line, it turns around and flies back to the runner. At this point the bird has traveled a distance of 1 . 32 km. Even though the bird is a dodo, we will assume that it occupies only one point in space, i.e. , a “zero” length bird. After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. L v b v r finish line How far does the bird travel? 1. 1.30 km ≤ d b < 1.40 km 2. 5.00 km ≤ d b ≤ ∞ km 3. 1.40 km ≤ d b < 1.50 km 4. 3.00 km ≤ d b < 5.00 km 5. 2.25 km ≤ d b < 2.50 km 6. 1.75 km ≤ d b < 2.00 km 7. 1.50 km ≤ d b < 1.75 km 8. 2.50 km ≤ d b < 2.75 km correct 9. 2.75 km ≤ d b < 3.00 km 10. 2.00 km ≤ d b < 2.25 km Explanation: Let : v r = 1 km / hr , L = 0 . 88 km , v b = 3 km / hr , and v b 1 = 1 . 32 km ....
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20 - fryer(vdf96 – Kinematics – graves –(6 1 This...

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