# h3 - Al Ali Ahmed Homework 3 Due 11:00 pm Inst Dr Mike...

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Al Ali, Ahmed – Homework 3 – Due: Sep 10 2007, 11:00 pm – Inst: Dr Mike Stokes 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. This is ONLINE HOMEWORK No 3 11.00 pm in Texas is 8.00 am the next morn- ing in Abu Dhabi 001 (part 1 oF 1) 10 points A rocket sled For testing equipment under large accelerations starts at rest and acceler- ates according to the expression a = (1 . 9 m / s 3 ) t + (2 . 7 m / s 2 ) . How Far does the sled move in the time interval t = 0 to t = 2 . 7 s? Correct answer: 16 . 0745 m. Explanation: Let : α = 1 . 9 m / s 3 , β = 2 . 7 m / s 2 , and t = 2 . 7 s . Basic Concepts: a = d v dt = d dt µ d x dt Solution: We are given a = α t + β , thus v = Z t 0 ( α t 0 + β ) dt 0 = α 2 t 2 + β t = d x dt x = Z t 0 α 2 t 0 2 + β t 0 · dt 0 = α 6 t 3 + β 2 t 2 = (1 . 9 m / s 3 ) 6 (2 . 7 s) 3 + (2 . 7 m / s 2 ) 2 (2 . 7 s) 2 = 16 . 0745 m . keywords: 002 (part 1 oF 2) 5 points A truck travels beneath an airplane that is moving 130 km / h at an angle oF 29 to the ground. a) How Fast must the truck travel to stay beneath the airplane? Correct answer: 113 . 701 km / h. Explanation: v y v x 130 km / h 29 Note: ±igure is not drawn to scale. Basic Concept: v x = v (cos θ ) Given: v = 130 km / h θ = 29 Solution: We need the plane’s horizontal component: v x = (130 km / h)(cos29 ) = 113 . 701 km / h 003 (part 2 oF 2) 5 points b) What is the magnitude oF the vertical com- ponent oF the velocity oF the plane? Correct answer: 63 . 0253 km / h. Explanation: Basic Concept: v y = v (sin θ ) Solution: v y = (130 km / h)(sin29 ) = 63 . 0253 km / h keywords: 004 (part 1 oF 2) 5 points Tammy leaves the o²ce, drives 31 km due north, then turns onto a second highway and

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Al Ali, Ahmed – Homework 3 – Due: Sep 10 2007, 11:00 pm – Inst: Dr Mike Stokes 2 continues in a direction of 31 north of east for 94 km. What is her total displacement from the oFce? Correct answer: 113 . 131 km. Explanation: Basic Concepts Choose a coordinate system with the positive x -axis representing 0 and the positive y -axis representing 90 . Solution S 1 S S S 2 S 1 S 2 S 2n S n S 2e S e θ α ±or the vector s 2 angled at θ north of east, the north component s 2 n is the side opposite the angle θ and the east component s 2 e is the side adjacent. The north component of the ²nal displace- ment is s n = s 1 + s 2 n = s 1 + s 2 sin θ The east component of the ²nal displacement is s e = s 2 e = s 2 cos θ The total displacement is the hypotenuse of the right triangle formed by s n and s e , so s = q s 2 n + s 2 e 005 (part 2 of 2) 5 points At what angle is her displacement? (Consider east to be 0 and north 90 .) Correct answer: 44 . 5846 . Explanation:
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## This note was uploaded on 02/16/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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h3 - Al Ali Ahmed Homework 3 Due 11:00 pm Inst Dr Mike...

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