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Unformatted text preview: Homework Set #3 September 27, 2005 1 Chapter 4 1.1 Problem 4.2 (a) The equation for x describes the motion in the i direction and the y equation describes the j direction. ~ r ( t ) = (18 . m/s ) t i + h (4 . 00 m/s ) t 4 . 90 m/s 2 t 2 i j (b) d~ r ( t ) dt = ~v ( t ) = (18 . m/s ) i + h (4 . 00 m/s ) 9 . 80 m/s 2 t i j (c) d~v ( t ) dt = ~a ( t ) = 9 . 80 m/s 2 j (d) Now just plug in the time of 3.00 s into the appropriate parts. ~ r (3 . 00 s ) = (18 . m/s ) (3 . 00 s ) i + (4 . 00 m/s ) (3 . 00 s ) 4 . 90 m/s 2 (3 . 00 s ) 2 j ~ r (3 . 00 s ) = (54 . m ) i (32 . 1 m ) j (e) ~v (3 . 00 s ) = (18 . m/s ) i + (4 . 00 m/s ) 9 . 8 m/s 2 (3 . 00) j ~v (3 . 00 s ) = (18 . m/s ) i (25 . 4 m/s ) j (f) ~a (3 . 00 s ) = 9 . 80 m/s 2 j 1.2 Problem 4.8 (a) Notice that the initial velocity is only in the i direction and the acceleration is only in the j direction. Now recall this equation of motion x ( t ) = x + v t + 1 2 at 2 Using this we can construct the equation of motion in the i direction x ( t ) = (500 m/s ) t and in the j direction y ( t ) = 1 2 3 . 00 m/s 2 t 2 = 1 . 50 m/s 2 t 2 1 Now we can combine these just as we did for problem 2. ~ r ( t ) = (500 m/s ) t i + 1 . 50 m/s 2 t 2 j Again to get the velocity take the time derivative of the position to get ~v ( t ) = (500 m/s ) i + 3 . 00 m/s 2 t j (b) So now we use t = 2 . 00 s to nd the position and velocity at that time. ~ r (2 . 00 s ) = (500 m/s ) (2 . 00 s ) i + 1 . 50 m/s 2 (2 . 00 s ) 2 j ~ r (2 . 00 s ) = 1 . 00 10 3 m i + 6 . 00 m j x = 1 . 00 10 3 m and y = 6 . 00 m ~v (2 . 00 s ) = (500 m/s ) i + 3 . 00 m/s 2 (2 . 00 s ) j ~v (2 . 00 s ) = (500 m/s ) i + (6 . 00 m/s ) j However, the problem asks for the speed not the velocity. v = q (500 m/s ) 2 + (6 . 00 m/s ) 2 = 500 m/s 1.3 Problem 4.10 This is projectile motion. Finding the x coordinate is simpler because there is no acceleration in this direction. This means that the x component of the velocity times the time is the x coordinate x = v x t = vcos ( ) t = (300 m/s ) cos (55 . 0) (42 . s ) = 7 .....
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 Fall '06
 Winnokur
 Physics, Work

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