This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Overton, Mays – Homework 3 – Due: Jan 28 2005, 4:00 am – Inst: Turner 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (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 8 s per grid square and on the vertical axis 8 m / s per grid square. 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 v ( t ) time × (8 s) velocity × (8m / s) Initially, at t = 0 the particle is at x = 62 m. What is the position x of the particle at time t = 32 s? Correct answer: 702 m. Explanation: Looking at the v ( t ) plot we see that over time t = 4 × 8 s = 32 s, the particle’s velocity decreases from the initial v = 4 × 8 m / s = 32 m / s to final v f = 1 × 8 m / s = 8 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 = 20 m / s . Consequently, the particle’s displacement during this time is simply Δ x = t × ¯ v = 640 m , and its final position x = x + Δ x = 702 m . 002 (part 2 of 2) 10 points What is the particle’s acceleration? Correct answer: . 75 m / s 2 . Explanation: The average acceleration of the particle is ¯ a = Δ v Δ t = v f v t = . 75 m / s 2 . Since the v ( t ) line is straight, the acceleration is constant, hence a = ¯ a = . 75 m / s 2 . 003 (part 1 of 1) 10 points A car traveling in a straight line has a velocity of 6 . 62 m / s at some instant. After 7 . 65 s, its velocity is 12 . 4 m / s. What is its average acceleration in this time interval? Correct answer: 0 . 755556 m / s 2 . Explanation: The average acceleration is a av = Δ v Δ t = v f v i Δ t = 12 . 4 m / s 6 . 62 m / s 7 . 65 s = 0 . 755556 m / s 2 004 (part 1 of 1) 10 points Vectors ~ A , ~ B , and ~ C are shown in the figure below. For convenience, the tails of each vector are arbitrarily located at (0,0)....
View
Full Document
 Spring '08
 Turner
 Physics, Acceleration, Work, Correct Answer, Overton, Mays

Click to edit the document details