The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.a(t) = 2t + 2, v(0) = −15, 0 ≤ t ≤ 5(a) Find the velocity at time t.v(t) = m/s(b) Find the distance traveled during the given time interval.
Question
The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.a(t) = 2t + 2, v(0) = −15, 0 ≤ t ≤ 5(a) Find the velocity at time t.v(t) = m/s(b) Find the distance traveled during the given time interval.
Solution
(a) To find the velocity at time t, we need to integrate the acceleration function a(t) with respect to time.
The integral of a(t) = 2t + 2 is ∫(2t + 2) dt = t^2 + 2t + C, where C is the constant of integration.
We know that the initial velocity v(0) = -15, so we can find C by setting t = 0 in the equation:
-15 = 0^2 + 2*0 + C -15 = C
So, the velocity function v(t) = t^2 + 2t - 15 m/s.
(b) To find the distance traveled, we need to find the displacement (which is the integral of the velocity function) and take the absolute value (since distance is always positive).
The integral of v(t) = t^2 + 2t - 15 is ∫(t^2 + 2t - 15) dt = (1/3)t^3 + t^2 - 15t + D, where D is another constant of integration.
We know that the particle starts at position 0, so D = 0.
So, the displacement function is s(t) = (1/3)t^3 + t^2 - 15t.
The displacement at t = 5 is s(5) = (1/3)5^3 + 5^2 - 155 = 125/3 + 25 - 75 = 50/3 m.
The displacement at t = 0 is s(0) = 0.
So, the distance traveled is |s(5) - s(0)| = |50/3 - 0| = 50/3 m.
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