# 2 - homework 02 – JACOBS AMBER – Due 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

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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....
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## 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.

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2 - homework 02 – JACOBS AMBER – Due 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

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