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HW #2 Solutions

# HW #2 Solutions - homework 02 FONTENOT BRIAN Due 4:00 am...

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homework 02 – FONTENOT, BRIAN – Due: Jan 28 2008, 4:00 am 1 Question 1, chap 2, sect 5. part 1 of 3 10 points In deep space (no gravity), the bolt (arrow) of a crossbow accelerates at 129 m / s 2 and attains a speed of 125 m / s when it leaves the bow. For how long is it accelerated? Correct answer: 0 . 968992 s (tolerance ± 1 %). Explanation: v = v o + at = at since the initial speed v o = 0, so t = v a = 125 m / s 129 m / s 2 = 0 . 968992 s Question 2, chap 2, sect 5. part 2 of 3 10 points What speed will the bolt have attained 4 . 5 s after leaving the crossbow? Correct answer: 125 m / s (tolerance ± 1 %). Explanation: It has left the crossbow, so a = 0 and the velocity of the bolt after 4 . 5 s will remain 125 m / s. Question 3, chap 2, sect 5. part 3 of 3 10 points How far will the bolt have traveled during the 4 . 5 s? Correct answer: 562 . 5 m (tolerance ± 1 %). Explanation: s = s o + v o t + 1 2 at 2 = vt 2 since s o = 0, v o = v , a = 0, and t = t 2 , so s = (125 m / s)(4 . 5 s) = 562 . 5 m Question 4, chap 2, sect 5. part 1 of 2 10 points A speed boat moving at 30 . 2 m / s ap- proaches a no-wake buoy marker 75 . 4 m ahead. The pilot slows the boat with a con- stant acceleration of - 4 . 2 m / s 2 by reducing the throttle. How long does it take the boat to reach the buoy? Correct answer: 3 . 21578 s (tolerance ± 1 %). Explanation: Given : v i = 30 . 2 m / s Δ x = 75 . 4 m a = - 4 . 2 m / s 2 Under the deceleration, Δ x = v i t + 1 2 a t 2 a t 2 + 2 v i t - 2 Δ x = 0 so that t = - 2 v i ± radicalbig 4 v i 2 + 8 a x ) 2 a = - 2(30 . 2) ± radicalBig 4(30 . 2) 2 + 8( - 4 . 2)(75 . 4) 2 ( - 4 . 2) 11 . 1652 , 3 . 21578 . The larger solution of 11 . 1652 is the second time the boat would pass the buoy (mov- ing backwards), assuming it could maintain a constant deceleration after it had stopped. Question 5, chap 2, sect 5. part 2 of 2 10 points What is the velocity of the boat when it reaches the buoy? Correct answer: 16 . 6937 m / s (tolerance ± 1 %). Explanation:

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homework 02 – FONTENOT, BRIAN – Due: Jan 28 2008, 4:00 am 2 v f = v i + a t = 30 . 2 m / s + ( - 4 . 2 m / s 2 ) (3 . 21578 s) = 16 . 6937 m / s . Question 6, chap 2, sect 6. part 1 of 1 10 points A ball is thrown straight up from the ground with an initial velocity of 37 . 2 m / s; at the same instant, a ball is dropped from the roof of a building 20 . 8 m high. After how long will the balls be at the same height? Correct answer: 0 . 55914 s (tolerance ± 1 %). Explanation: Basic Concepts: In the case of each ball the expression d = d 0 + v 0 t - 1 2 gt 2 is adequate to describe the situation. Solution: Since the falling ball has no initial velocity we obtain d 1 = h - 1 2 gt 2 For the second ball, r 0 > 0 and d 0 = 0, so d 2 = v 0 t - 1 2 gt 2 . To find the time when the two balls are at equal height from the ground, we need to equate the two equations that we have devel- oped: h - 1 2 gt 2 = v 0 t - 1 2 gt 2 which simplifies to h = v 0 t and we solve for t. Question 7, chap 2, sect 5. part 1 of 1 10 points A baby sitter pushing a stroller starts from rest and accelerates uniformly at a rate of 0.705 m/s 2 .
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