This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: homework 02 – JACOBS, AMBER – Due: Jan 29 2008, 1:00 am 1 Question 1, chap 2, sect 8. part 1 of 1 10 points The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given by a = α v 2 , where v > 0 m/s and α = − 4 . 8 m − 1 . If the marble enters this fluid with a speed of 2 . 11 m / s, how long will it take before the marble’s speed is reduced to half of its initial value? Correct answer: 0 . 0987362 s (tolerance ± 1 %). Explanation: Basic Concept: a = d v dt Solution: a = d v dt = α v 2 Separating variables, dv v 2 = α dt . Integrating this, we have integraldisplay dv v 2 = α integraldisplay dt. With our limits of integration, this becomes integraldisplay v v = v u − 2 du = α integraldisplay t t =0 dt ′ − 1 v + 1 v = α t. We want to know t for which v = v / 2: t = 1 α parenleftbigg 1 v − 2 v parenrightbigg = − 1 α v = − 1 ( − 4 . 8 m − 1 )(2 . 11 m / s) = 0 . 0987362 s . Question 2, chap 2, sect 8. part 1 of 1 10 points The velocity of a particle moving along the x axis is given by v x = a t − b t 3 for t > , where a = 32 m / s 2 , b = 2 . 9 m / s 4 , and t is in s. What is the acceleration a x of the particle when it achieves its maximum displacement in the positive x direction? Correct answer: − 64 m / s 2 (tolerance ± 1 %). Explanation: The time when it achieves maximum dis placement is when its velocity is 0 v = a t − b t 3 = 0 t ( a − b t 2 ) = 0 . Solve for t , we get t = radicalbigg a b = radicalBigg 32 m / s 2 2 . 9 m / s 4 = 3 . 32182 s . The acceleration at this particular t = radicalbigg a b is then given by a x = d v x dt = d ( a t − b t 3 ) dt = a − 3 b t 2 = a − 3 a = − 2 a = − 2 (32 m / s 2 ) = − 64 m / s 2 . Question 3, chap 2, sect 8. part 1 of 3 10 points The position of a softball tossed vertically upward is described by the equation y = c 1 t − c 2 t 2 , where y is in meters, t in seconds, c 1 = 3 . 56 m / s, and c 2 = 6 . 06 m / s 2 . Find the ball’s initial speed v at t = 0 s. homework 02 – JACOBS, AMBER – Due: Jan 29 2008, 1:00 am 2 Correct answer: 3 . 56 m / s (tolerance ± 1 %). Explanation: Basic Concepts: v = dx dt a = dv dt = d 2 x dt 2 Solution: The velocity is simply the derivative of y with respect to t : v = dy dt = 3 . 56 m / s − 2(6 . 06 m / s 2 ) t, which at t = 0 is v = 3 . 56 m / s . Question 4, chap 2, sect 8. part 2 of 3 10 points Find its velocity at t = 1 . 53 s. Correct answer: − 14 . 9836 m / s (tolerance ± 1 %). Explanation: Substituting t = 1 . 53 s into the above for mula for v , we obtain v = 3 . 56 m / s − 2(6 . 06 m / s 2 )(1 . 53 s) = − 14 . 9836 m / s . Question 5, chap 2, sect 8....
View
Full
Document
This note was uploaded on 04/28/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas at Austin.
 Spring '07
 KOPP
 Physics, Acceleration, Work

Click to edit the document details