An object has a constant acceleration of 40 ft/sec2, an initial velocity of - 20 ft/sec, and an initial position of 10 ft. Find the position function, s (t), describing the motion of the object.

I'm all the way up to here, what do I do now?

a (t) = 40 v (0) = - 20 s (0) = 10

the integral of 40 dt, antiderivative is v (t) = 40t+C

40 (0) + C=-20

C=-60

s (t) = integral of (40 (t) - 60) dt

antiderivative is 20t^2-60t+C

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28 January, 06:01

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Answers (1)

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Damon Mathis · Answer 1

D2x/dt2=40

dx/dt=40t+vo and vo=-20 so

dx/dt=40t-20

x (t) = 40t^2/2-20t+x0 and x0=10 so

x (t) = 20t^2-20t+10