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Unformatted text preview: homework 03 FIERRO, JEFFREY Due: Jan 25 2008, 11:00 pm 1 Question 1, chap 2, sect 5. part 1 of 1 10 points An electron, starting from rest and moving with a constant acceleration, travels 4 . 6 cm in 13 ms. What is the magnitude of this acceleration in km/s 2 ? Correct answer: 0 . 544379 km / s 2 (tolerance 1 %). Explanation: An electron starting from rest with con stant acceleration will travel a distance of d = 1 2 at 2 in t seconds. Given the time and distance traveled we solve for a : a = 2 d t 2 making any necessary unit conversions along the way. Question 2, chap 2, sect 7. 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 2 . 12 m / s, how long will it take before the marbles speed is reduced to half of its initial value? Correct answer: 0 . 0982704 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 )(2 . 12 m / s) = 0 . 0982704 s . Question 3, chap 2, sect 6. part 1 of 2 10 points A ball is thrown upward. After reaching a maximum height, it continues falling back towards Earth. On the way down, the ball is caught at the same height at which it was thrown upward. Neglect: Air resistance. Its initial vertical speed v , acceleration of gravity is 9 . 8 m / s 2 , and maximum height h max are shown in the figure below. b b b b b b b b b bb b b b b b b b b b v 9 . 8m / s 2 h max homework 03 FIERRO, JEFFREY Due: Jan 25 2008, 11:00 pm 2 If the time ( up and down ) the ball remains in the air is 2 . 28 s, calculate its speed when it caught. Correct answer: 11 . 172 m / s (tolerance 1 %). Explanation: Basic Concept: For constant accelera tion, we have v = g t (1) y = y + v t + 1 2 a t 2 . (2) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing h max = y y , a , and t . Choose the positive direction to be up, then a = g and t up = v g ....
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 Spring '08
 Turner
 Acceleration, Work

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