02_InstSolManual_PC - MOTION IN A STRAIGHT LINE 2 x t...

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MOTION IN A STRAIGHT LINE E XERCISES Section 2.1 Average Motion 13. I NTERPRET We need to find average speed, given distance and time. D EVELOP Speed is distance divided by time. E VALUATE 100 m 9.77 s 10.2 m/s v = = A SSESS His time is about 10 seconds, so his speed is about 10 m/s. 14. I NTERPRET We need to find average speed, given distance and time. D EVELOP The Olympic marathon distance is 42,186 m. We convert her time to seconds, and use x t v = to find her speed. E VALUATE 3600 s 60 s 1 h 1 min 2 h 26 min 20 s (2 h) (26 min) (20 s) 8780 s. + + = × + × + = 42,186 m 4.805 m/s 8780 s x v t = = = A SSESS To check our answer we can convert this speed to mph, and see if it’s reasonable for a distance of about 26 miles in about 2.5 hours. 15. I NTERPRET This is a one-dimensional kinematics problem, and we identify the bicyclist as the object of interest. His trip consists of two parts (out and back), and we are asked to compute the displacement and average velocity of each part, as well as for the trip as a whole. D EVELOP For motion in a straight line, the displacement, or the net change in position, is 2 1 , x x x = - where x 1 and x 2 are the starting and the end points, respectively. The average velocity is the displacement divided by the time interval, / , v x t = ∆ ∆ as shown in Equation 2.1. In our coordinate system, we take north to be + x . E VALUATE (a) The displacement at the end of the first 2.5 h is out 2 1 24 km 0 km 24 km x x x = - = - = (b) With out 2.5 h, t = the average velocity over this interval is out out out 24 km 9.6 km h (north) 2.5 h x v t = = = / (c) With back 1.5 h, t = the average velocity for the homeward leg of the trip is back back back 24 km 16 km/h (south) 1.5 h wx v t - = = = - (d) Since the final position of the bicyclist is the same as his initial position, his total displacement of the trip is total out back 24 km ( 24 km) 0 km. x x x = ∆ + ∆ = + - = (e) Since total 0, x = the average velocity for the entire trip is total total total / 0. v x t = ∆ = A SSESS Note the distinction between average velocity and average speed. The former depends only on the net displacement, while the latter takes into consideration the total distance traveled. The average speed for this trip is out back out back | | | | 24 km 24 km average speed 12 km/h 2.5 h 1.5 h x x t t + ∆ + = = = + ∆ + 2.1 2
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2.2 Chapter 2 16. I NTERPRET We must find the time it takes for a radio signal to travel a given distance. D EVELOP Radio waves travel at a constant speed of 8 3.0 1 0 v = × m/s. We know that the distance is 9 12 1 .2 1 0 km 1.2 1 0 m, x = × = × so we can use the definition of speed, , x t v = to calculate the time. E
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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02_InstSolManual_PC - MOTION IN A STRAIGHT LINE 2 x t...

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