The acceleration function in (m/s^2) is given as a(t)= 5-2t and the initial velocity is given as v(0)= -6 for a particle moving along a line.

Question

A) find the v(t)

B) find the distance traveled during the time interval 1 ≤ t ≤ 3

Answer 1

A. We know acceleration is the derivative of velocity, thus velocity must be the antiderivative of acceleration.

Integrating a(t) = 5 - 2t leads us to v(t) = 5t - t^2 + C (you may derive back to a(t) to reassure yourself)

To determine what our constant, C, is, we must use the fact v(0) = -6.

v(0) = -6 = 5(0) - 0^2 + C. Simply put, -6 = C. So, v(t), specifically, is 5t - t^2 - 6

B. Similarly, velocity is the derivative of position, so we need to integrate v(t) to determine position.

This time, however, it is a definite integral over the span of t = 1 to t = 3.

Our integral comes out to p(t) = (5/2)t^2 - t^3/3 - 6t + C

Now, we plug in t = 3 and subtract the result of t = 1 from it.

p(3) = (5/2)(3)^2 - (3)^3 / 3 - 6(3) = (5/2)(9) - 27 / 3 - 18 = 22.5 - 9 - 18 = -4.5

p(1) = (5/2)(1)^2 - (1)^3 / 3 - 6(1) = (5/2) - (1/3) - 6 = -3.83

p(3) - p(1) = -0.67, so the particle moved a distance of 0.67 meters in the opposite direction.

Hope this helped!

Answer 2

v(t) = -6 + ∫₀^t (5-2t) dt = -6 + 5t - t²

^{The distance is an integral of the absolute value of V}