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Unformatted text preview: Overton, Mays – Homework 5 – Due: Feb 3 2005, 4:00 am – Inst: Turner 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A speeder traveling at a constant speed of 85 km / h races past a billboard. A patrol car pursues from rest with constant acceleration of 5 (km / h) / s until it reaches its maximum speed of 129 km / h, which it maintains until it catches up with the speeder. How long does it take the patrol car to catch the speeder if it starts moving just as the speeder passes? Correct answer: 37 . 8205 s. Explanation: Let : v s = 85 km / h , a = 5 (km / h) / s , and v max = 129 km / h . The patrol car takes t 1 seconds to reach its maximum speed, which it maintains for t 2 seconds until it catches the speeder. During this time, the speeder’s constant velocity is v s = (85 km / h) ‡ 1000 m 1 km ·‡ 1 h 3600 s · = 23 . 6111 m / s and its final position is Δ x s = v s ( t 1 + t 2 ) . While the patrol car is accelerating, v max = (129 km / h) ‡ 1000 m 1 km ·‡ 1 h 3600 s · = 35 . 8333 m / s Applying the kinematics equations, v max = v + at 1 = at 1 t 1 = v max a = 129 km / h 5 (km / h) / s = 25 . 8 s and its displacement is x 1 = v t 1 + 1 2 at 2 1 = 1 2 at 2 1 = [5 (km / h) / s](25 . 8 s) 2 2 × 1000 m 1 km × 1 h 3600 s = 462 . 25 m The final position of the patrol car is Δ x p = x 1 + v max t 2 . The displacements are the same, so Δ x p = Δ x s x 1 + v max t 2 = v s ( t 1 + t 2 ) t 2 ( v max v s ) = v s t 1 x 1 Since v s t 1 = (23 . 6111 m / s)(25 . 8 s) = 609 . 167 m , we have t 2 = v s t 1 x 1 v max v s = (609 . 167 m) (462 . 25 m) (35 . 8333 m / s) (23 . 6111 m / s) = 12 . 0205 s , and the total time is t = t 1 + t 2 = 25 . 8 s + 12 . 0205 s = 37 . 8205 s . 002 (part 2 of 2) 10 points How far does each car travel? Correct answer: 892 . 983 m. Explanation: Using the speeder’s information, Δ x = v s Δ t = (23 . 6111 m / s)(37 . 8205 s) = 892 . 983 m . Overton, Mays – Homework 5 – Due: Feb 3 2005, 4:00 am – Inst: Turner 2 10 20 30 40 200 400 600 800 1000 1200 1400 x (m) t (s) Speeder Officer 003 (part 1 of 1) 10 points Assume: The speed of sound in air is 317 m / s....
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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.
 Spring '08
 Turner
 Physics, Work

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