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PHY303K HW1

# PHY303K HW1 - perez(nap563 HW01 Tsoi(58160 This print-out...

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perez (nap563) – HW01 – Tsoi – (58160) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below. P x t At point P, the mass has 1. positive velocity and negative accelera- tion. correct 2. positive velocity and positive accelera- tion. 3. negative velocity and negative accelera- tion. 4. negative velocity and positive accelera- tion. 5. zero velocity and zero acceleration. 6. negative velocity and zero acceleration. 7. zero velocity but is accelerating (posi- tively or negatively). 8. positive velocity and zero acceleration. Explanation: The velocity is positive because the slope of the curve at P is positive. The acceleration is negative because the curve is concave down at P. 002 10.0 points Consider three position curves between time points t A and t B . s B s A B A t A t B 3 2 1 0 t s Choose the correct relationship among quantities v 1 , v 2 , and v 3 . v = v A + v B 2 when a is a constant. 1. v 1 > v 2 > v 3 2. v 1 < v 2 < v 3 3. v 1 = v 2 = v 3 correct Explanation: The average velocity of an object is v = displacement time = s B - s A t B - t A . All three curves have exactly the same change in position Δ s = s B - s A in exactly the same time interval Δ t = t B - t A , so all three average velocities are equal: v 1 = v 2 = v 3 . 003 (part 1 of 2) 10.0 points Consider the position-time graph for a squir- rel running along a clothesline. 1 2 3 4 5 0 1 2 3 4 - 1 - 2 time (s) position (m) What is the squirrel’s displacement at the time t = 4 . 0 s? Your answer must be within ± 5.0% Correct answer: - 1 m. Explanation:

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perez (nap563) – HW01 – Tsoi – (58160) 2 Let : x f = - 1 m and x i = 0 m . The displacement is Δ x = x f - x i = ( - 1 m) - (0 m) = - 1 m . 004 (part 2 of 2) 10.0 points What is the squirrel’s average velocity during the time interval between 0.0 s and 4.0 s? Your answer must be within ± 5.0% Correct answer: - 0 . 25 m / s. Explanation: Let : Δ t = 4 . 0 s . v avg = Δ x Δ t = - 1 m 4 s = - 0 . 25 m / s . 005 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 0 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 0 t Q t R t S t P 2. None of these graphs is correct. 3. t x 0 t Q t R t S t P 4. t x 0 t Q t R t S t P 5. t x 0 t Q t R t S t P 6. t x 0 t Q t R t S t P 7. t x 0 t Q t R t S t P correct 8. t x 0 t Q t R t S t P 9. t x 0 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.
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