A particle moves along the x axis with velocity vx = (9.5 m/s2)t - 6.6 m/s. (a)

Question

A particle moves along the x axis with velocity vx = (9.5 m/s2)t - 6.6 m/s.
(a) Find the average acceleration for two different one-second intervals, one beginning at t = 3.0 s and the other beginning at t = 4.0 s.
aavg, 3.0 to 4.0 s_________m/s^2
aavg, 4.0 to 5.0 s______m/s^2

(b) Sketch vx versus t over the interval 0 < t < 10 s. (Do this on paper. Your instructor may ask you to turn in this work.)

(c) How do the instantaneous accelerations at the middle of each of the two time intervals specified in Part (a) compare to the average accelerations found in Part (a)?

Explanation / Answer

average accn = final velocity - initial velocity / time taken for the change. = a= 9.5 m/s^2 v at t= 3 ==> 21.9 v at t= 4 ==> 31.4 v at t= 5 ==> 40.9 aavg, 3.0 to 4.0 s== 9.5 m/s^2 aavg, 4.0 to 5.0 == 9.5 m/s^2 instantaneous accn = dv/dt = 9.5m/s^2 they are equal.

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