31_PartUniversity Physics Solution

31_PartUniversity Physics Solution - 2 MOTION ALONG A...

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2-1 M OTION A LONG A S TRAIGHT L INE 2.1. IDENTIFY: The average velocity is av- x x v t Δ = Δ . SET UP: Let x + be upward. EXECUTE: (a) av- 1000 m 63 m 197 m/s 4.75 s x v == (b) av- 1000 m 0 169 m/s 5.90 s x v EVALUATE: For the first 1.15 s of the flight, av- 63 m 0 54.8 m/s 1.15 s x v . When the velocity isn&t constant the average velocity depends on the time interval chosen. In this motion the velocity is increasing. 2.2. IDENTIFY: av- x x v t Δ = Δ SET UP: 5 13.5 days 1.166 10 s . At the release point, 6 5.150 10 m x =+ × . EXECUTE: (a) 6 21 av- 6 4.42 m/s x xx v t −× = Δ× (b) For the round trip, x x = and 0 x Δ= . The average velocity is zero. EVALUATE: The average velocity for the trip from the nest to the release point is positive. 2.3. IDENTIFY: Target variable is the time t Δ it takes to make the trip in heavy traffic. Use Eq.(2.2) that relates the average velocity to the displacement and average time. SET UP: av- x x v t Δ = Δ so av- x x vt Δ and av- . x x t v Δ EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two cities: av- (105 km/h)(1 h/60 min)(140 min) 245 km. x xv t = Now use av- x v for heavy traffic to calculate ; t Δ x Δ is the same as before: av- 245 km 3.50 h 3 h 70 km/h x x t v Δ = = = and 30 min. The trip takes an additional 1 hour and 10 minutes. EVALUATE: The time is inversely proportional to the average speed, so the time in traffic is (105/70)(140 m) 210 min. = 2.4. IDENTIFY: The average velocity is av- x x v t Δ = Δ . Use the average speed for each segment to find the time traveled in that segment. The average speed is the distance traveled by the time. SET UP: The post is 80 m west of the pillar. The total distance traveled is 200 m 280 m 480 m + = . EXECUTE: (a) The eastward run takes time 200 m 40.0 s 5.0 m/s = and the westward run takes 280 m 70.0 s 4.0 m/s = . The average speed for the entire trip is 480 m 4.4 m/s 110.0 s = . (b) av- 80 m 0.73 m/s 110.0 s x x v t Δ− = Δ . The average velocity is directed westward. 2

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2-2 Chapter 2 EVALUATE: The displacement is much less than the distance traveled and the magnitude of the average velocity is much less than the average speed. The average speed for the entire trip has a value that lies between the average speed for the two segments. 2.5. IDENTIFY: When they first meet the sum of the distances they have run is 200 m. SET UP: Each runs with constant speed and continues around the track in the same direction, so the distance each runs is given by dv t = . Let the two runners be objects A and B .
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31_PartUniversity Physics Solution - 2 MOTION ALONG A...

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