HW #2 Solutions - homework 02 FONTENOT, BRIAN Due: Jan 28...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 x = 75 . 4 m a =- 4 . 2 m / s 2 Under the deceleration, x = v i t + 1 2 a t 2 a t 2 + 2 v i t- 2 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: 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 + v 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 and d = 0, so d 2 = v t- 1 2 gt 2 ....
View Full Document

This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 9

HW #2 Solutions - homework 02 FONTENOT, BRIAN Due: Jan 28...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online