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Unformatted text preview: homework 02 ALLEN, HUNTER 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 = v 2 , where v > 0 m/s and = 4 . 8 m 1 . If the marble enters this fluid with a speed of 1 . 07 m / s, how long will it take before the marbles speed is reduced to half of its initial value? Correct answer: 0 . 194704 s (tolerance 1 %). Explanation: Basic Concept: a = d v dt Solution: a = d v dt = v 2 Separating variables, dv v 2 = dt . Integrating this, we have integraldisplay dv v 2 = integraldisplay dt. With our limits of integration, this becomes integraldisplay v v = v u 2 du = integraldisplay t t =0 dt 1 v + 1 v = t. We want to know t for which v = v / 2: t = 1 parenleftbigg 1 v 2 v parenrightbigg = 1 v = 1 ( 4 . 8 m 1 )(1 . 07 m / s) = 0 . 194704 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 = 31 m / s 2 , b = 1 . 7 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: 62 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 31 m / s 2 1 . 7 m / s 4 = 4 . 27028 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 (31 m / s 2 ) = 62 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 = 7 . 58 m / s, and c 2 = 3 . 15 m / s 2 . Find the balls initial speed v at t = 0 s. homework 02 ALLEN, HUNTER Due: Jan 29 2008, 1:00 am 2 Correct answer: 7 . 58 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 = 7 . 58 m / s 2(3 . 15 m / s 2 ) t, which at t = 0 is v = 7 . 58 m / s . Question 4, chap 2, sect 8. part 2 of 3 10 points Find its velocity at t = 1 . 96 s. Correct answer: 4 . 768 m / s (tolerance 1 %). Explanation: Substituting t = 1 . 96 s into the above for mula for v , we obtain v = 7 . 58 m / s 2(3 . 15 m / s 2 )(1 . 96 s) = 4 . 768 m / s . Question 5, chap 2, sect 8. part 3 of 3 10 points Find its acceleration at t = 1 . 96 s....
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 Spring '07
 KOPP
 Physics, Acceleration, Work

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