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Unformatted text preview: oldhomewk 04 – YOO, HEE – Due: Jan 24 2008, 4:00 am 1 Question 1, chap 2, sect 4. part 1 of 2 10 points The velocity v ( t ) of some particle is plotted as a function of time on the graph below. The scale on the horizontal axis is 7 s per division and on the vertical axis 6 m / s per division. 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 v ( t ) time ( × 7 s) velocity( × 6m / s) Initially, at t = 0 the particle is at x = 35 m. What is the position x of the particle at time t = 28 s? Correct answer: 455 m (tolerance ± 1 %). Explanation: Looking at the v ( t ) plot we see that over time t = 4 × 7 s = 28 s, the particle’s velocity decreases from the initial v = 4 × 6 m / s = 24 m / s to final v f = 1 × 6 m / s = 6 m / s . The v ( t ) line is straight, which indicates constant deceleration rate, hence the average velocity is given by ¯ v = v + v f 2 = 15 m / s . Consequently, the particle’s displacement during this time is simply Δ x = t × ¯ v = 420 m , and its final position x = x + Δ x = 455 m . Question 2, chap 2, sect 4. part 2 of 2 10 points What is the particle’s acceleration? Correct answer: . 642857 m / s 2 (tolerance ± 1 %). Explanation: The average acceleration of the particle is ¯ a = Δ v Δ t = v f v t = . 642857 m / s 2 . Since the v ( t ) line is straight, the acceleration is constant, hence a = ¯ a = . 642857 m / s 2 . Question 3, chap 2, sect 5. part 1 of 2 10 points A speeder passes a parked police car at 34.8 m/s. Instantaneously, the police car starts from rest with a uniform acceleration of 2.44 m/s 2 ....
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This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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