problem02_18 solution

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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2.18: a) The velocity at t = 0 is (3.00 s m ) + (0.100 3 s m ) (0) = 3.00 s m , and the velocity at t = 5.00 s is (3.00 s m ) + (0.100 3 s m ) (5.00 s) 2 = 5.50 s m , so Eq. (2.4) gives the average acceleration as 2 s m 50 . ) s 00 . 5 ( ) s m 00 . 3 ( ) s m 50 . 5 ( = - . b) The instantaneous acceleration is obtained by using Eq. (2.5),
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Unformatted text preview: . ) s m 2 . ( 2 3 t t dt dv a x = = = β Then, i) at t = 0, a x = (0.2 3 s m ) (0) = 0, and ii) at t = 5.00 s, a x = (0.2 3 s m ) (5.00 s) = 1.0 2 s m ....
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• Acceleration, ax, Average Acceleration, instantaneous acceleration, t. dt

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