Phys111-2sol

# Phys111-2sol - Anderson, Matthew Homework 2 Due: Sep 21...

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Unformatted text preview: Anderson, Matthew Homework 2 Due: Sep 21 2007, 11:00 pm Inst: Opyrchal, H 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A car starts from rest and accelerates uni- formly for 7 s. The distance traveled in the 7 s is greater than the distance traveled in the first second by a factor of: Correct answer: 49 . Explanation: The distance traveled during time t is 1 2 g t 2 . Therefore the ratio of the distances is given by 1 2 g (7 s) 2 1 2 g (1 s) 2 = 49 keywords: 002 (part 1 of 1) 10 points A car, moving along a straight stretch of high- way, begins to accelerate at 0 . 0455 m / s 2 . It takes the car 26 . 7 s to cover 1 km. How fast was the car going when it first began to accelerate? Correct answer: 36 . 8458 m / s. Explanation: We can describe this situation with the equation d = v t + 1 2 at 2 Given the time, distance and acceleration we simply solve for the initial velocity v = d t- 1 2 at converting the distance from km to m to ob- tain the proper units. keywords: 003 (part 1 of 1) 10 points A driver in a car traveling at a speed of 71 . 6 mi / h sees a deer 104 . 2 m away on the road. Assuming the deer does not move, what is the minimum constant acceleration necessary for the car to stop without hitting the deer? Correct answer:- 4 . 91522 m / s 2 . Explanation: Let : v i = 71 . 6 mi / h , v f = 0 mi / h , and x = 104 . 2 m . The acceleration can be found from v f 2 = v i 2 + 2 a ( x ) = 0 , since v f = 0 m / s....
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## This note was uploaded on 10/21/2010 for the course PHYS 003 taught by Professor Janow during the Fall '10 term at NJIT.

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Phys111-2sol - Anderson, Matthew Homework 2 Due: Sep 21...

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