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Unformatted text preview: P207  Fall 2008 Solutions Assignment 2 1 (a) For a constant acceleration a , and initial position and velocity zero, v ( t ) = at (1) and x ( t ) = 1 2 at 2 (2) Solve Equation 1 for time to get t = v/a (3) Then substitution of Equation 3 into Equation 2 gives x = 1 2 a parenleftBig v a parenrightBig 2 = 1 2 v 2 a (4) Finally solve Equation 4 for a . a = 1 2 v 2 x (5) In order to reach a velocity of 360 km/h in a distance of 1.8km, we have, using Equation 5 that the acceleration will be a = 1 2 360 2 1 . 8 (km / hr) 2 km = 3 . 6 10 4 km / hr 2 = 3 . 6 10 4 km / hr 2 parenleftbigg 1000 m / km (3600 s / hr) 2 parenrightbigg = 2 . 8m / s 2 = 2 . 8 m / s 2 9 . 8 m / s 2 = 0 . 29 g (b) From Equation 3 we see that it will take a time t = 360 km / hr 3 . 6 10 4 km / hr 2 = 10 2 hr = 36s 1 (c) Acceleration, velocity and displacement vs time are plotted below. 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 5 10 15 20 25 30 35 displacement [km] time [seconds] 50 100 150 200 250 300 350 400 5 10 15 20 25 30 35 velocity [km/hr] 0.5 1 1.5 2 2.5 3 3.5 4 5 10 15 20 25 30 35 acceleration [10 4 km/hr 2 ] (d) The average acceleration of a car is v t = 100 km / hr (9 s)(1 / 3600)hr / s = 4 10 4 km / hr 2 = 100 km / hr( . 001 m / km 3600 s / hr ) 9 s = 3 . 1 m / s 2 (as compared to 3 . 6 10 4 km / hr 2 = 2 . 8 m / s 2 for the jet) (e) Runway 26 appears to be about half the length of Runway 22 or x = 1 2 (2 . 1)km = 1 . 05 km. In order to determine the velocity after a uniform acceleration a for a distance x we solve Equation 5 for v and find that v = 2 ax = ( 2(3 . 6 10 4 km / hr 2 )(1 . 05 km) ) 1 / 2 = 275 km / hr 2. Chapter 2, Problem 38. The distance that the car travels before coming to a stop is x = v T + v T dec 1 2 aT 2 dec 2 where T dec is the time it takes to come to a stop from the moment you apply...
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 Fall '07
 LIEPE, M
 Physics, Acceleration

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