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Calculus help?

The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.

a(t) = 2t + 4, v(0) = −5, 0 ≤ t ≤ 5

(a) Find the velocity at time t.

Find v(t) in m/s

Then, find the distance traveled during the given time interval.(in meters)

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  • 2 months ago

    Acceleration is the rate of change of velocity, and so velocity is the antiderivative of acceleration.  If a(t) = 2t+4 , then v(t) = t^2 + 4t + C.  Knowing that v(0) = -5, we know that C is -5.  v(t) = t^2 + 4t -5.  .  velocity is the rate of change of position, so position is the antiderivative of velocity.  s(t) = t^3/3 +2t^2 -5t + C.  To get the distance you just need to take  s(5) - s(0) .  All the terms except C in s(0) are zero, and the C in s(5) is cancelled out by the C in s(0), so the answer is just  5^3/3 + 2(5^2) - 5(5) = 125/3 + 50 - 25 = 41.66666 + 25 = 66.66666

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