The position of an object as a function of time is given as x = At3 + Bt2 + Ct +

Question

The position of an object as a function of time is given as x
= At3 + Bt2 + Ct + D. The constants are A = 2.05 m/s3, B
= 1.23 m/s2, C = -4.21 m/s, and D = 3.97 m.

a) What is the velocity of the object at t = 19.9 s?

b) At what time is the object at rest? (Start at t = 0 s and
consider only positive time.)

c) What is the acceleration of the object at t = 0.447 s?

Explanation / Answer

vel = dx/dt = 3At^2 + 2Bt + C + 0 (a) vel at t=19.9s = 3 x 2.05 x 19.9^2 + 2 x 1.23 x 19.9 + (-4.21) m/s =2435.4615 + 48.954 - 4.21 =2480.2055 m/s ans. (b) object at rest ==> v=0 ==> 3 x 2.05 x t^2 + 2 x 1.23 t -4.21 =0 ===> t= 0.651 s. ans. (c) a = dv/dt =6At + 2B + 0 AT t = 0.447, a = 6 x 2.05 x 0.447 + 2 x 1.23 = 7.9581 m/s^2 ans.

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