Chap 2 - MOTION IN A STRAIGHT LINE 2 x t EXERCISES Section...

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2.1 MOTION IN A STRAIGHT LINE EXERCISES Section 2.1 Average Motion 13. INTERPRET We need to find average speed, given distance and time. DEVELOP Speed is distance divided by time. EVALUATE v . == 100 10 2 m 9.77 s m/s ASSESS His time is about 10 seconds, so his speed is about 10 m/s. 14. INTERPRET We need to find average speed, given distance and time. DEVELOP The Olympic marathon distance is 42,186 m. We convert her time to seconds, and use v x t = to find her speed. EVALUATE 22 6 2 0 2 2 6 3600 1 60 hs h s h ++ = × + × min ( ) ( min) s ss 1 20 8780 min () . += v x t = 42 186 4 805 , ./ m 8780 s ms ASSESS 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. INTERPRET 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. DEVELOP For motion in a straight line, the displacement, or the net change in position, is xx x =− 21 , 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, vx t =∆ ∆ /, as shown in Equation 2.1. In our coordinate system, we take north to be + x . EVALUATE (a) The displacement at the end of the first 2.5 h is xx x out km km km =−= = 24 0 24 (b) With t out h, = 25 . the average velocity over this interval is v x t out out out km h 9.6 km h (north) = / 24 . (c) With t back h, = 15 . the average velocity for the homeward leg of the trip is v wx t back back back km h 16 km/h (so 24 . uth) (d) Since the final position of the bicyclist is the same as his initial position, his total displacement of the trip is ∆∆ x total out back km km km. =+ = + = 24 24 0 (e) Since x total = 0, the average velocity for the entire trip is t total total total /. 0 2

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2.2 Chapter 2 ASSESS 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 average speed || | | out back out back = + + = ∆∆ xx tt 24 24 25 15 12 km km hh km h + + = .. / 16. INTERPRET We must find the time it takes for a radio signal to travel a given distance. DEVELOP Radio waves travel at a constant speed of v . 30 10 8 m/s. We know that the distance is x 12 91 2 km m, so we can use the definition of speed, v x t = , to calculate the time. EVALUATE vt x t x v =→== × × 12 10 30 10 12 8 . . meters meters secon d s. = 4000 We can convert this to a more conveniently sized unit if we wish: 4000 1 11 s 1 min 60 s h 60 min h ×× = . ASSESS This may seem like a long time, but the solar system is a big place. 17. INTERPRET We need to find average speed, given distance and time.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at Berkeley.

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Chap 2 - MOTION IN A STRAIGHT LINE 2 x t EXERCISES Section...

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