ch02-p020 - (f) A maximum v requires a = 0, which occurs...

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20. (a) Taking derivatives of x ( t ) = 12 t 2 – 2 t 3 we obtain the velocity and the acceleration functions: v ( t ) = 24 t – 6 t 2 and a ( t ) = 24 – 12 t with length in meters and time in seconds. Plugging in the value t = 3 yields (3) 54 m x = . (b) Similarly, plugging in the value t = 3 yields v (3) = 18 m/s. (c) For t = 3, a (3) = –12 m/s 2 . (d) At the maximum x , we must have v = 0; eliminating the t = 0 root, the velocity equation reveals t = 24/6 = 4 s for the time of maximum x . Plugging t = 4 into the equation for x leads to x = 64 m for the largest x value reached by the particle. (e) From (d), we see that the x reaches its maximum at t = 4.0 s.
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Unformatted text preview: (f) A maximum v requires a = 0, which occurs when t = 24/12 = 2.0 s. This, inserted into the velocity equation, gives v max = 24 m/s. (g) From (f), we see that the maximum of v occurs at t = 24/12 = 2.0 s. (h) In part (e), the particle was (momentarily) motionless at t = 4 s. The acceleration at that time is readily found to be 24 12(4) = 24 m/s 2 . (i) The average velocity is defined by Eq. 2-2, so we see that the values of x at t = 0 and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part (a)). Thus, v avg = 54 0 3 0 = 18 m/s ....
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