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Unformatted text preview: oldhomewk 04 NOORANI, ZOHEB 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 particles 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 particles displacement during this time is simply x = t v = 420 m , and its final position x = x + x = 455 m . Question 2, chap 2, sect 4. part 2 of 2 10 points What is the particles acceleration? Correct answer: . 642857 m / s 2 (tolerance 1 %). Explanation: The average acceleration of the particle is a = v 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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 Spring '08
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