Question

A particle moves along the x-axis with the given acceleration function a(t) = 8t - 6, initial position s(0) = 2, and initial velocity v(0). Find the position function.

Answer 1

Answer:

$s(t) = \frac{4t^3}{3} - 3t^2 + v(0)t + 2$

Step-by-step explanation:

Relation between acceleration, velocity and position:

The velocity function is the integral of the acceleration function.

The position function is the integral of the velocity function.

Acceleration:

As given by the problem, the acceleration function is:

$a(t) = 8t - 6$

Velocity:

$v(t) = \int a(t) dt = \int (8t - 6) dt = \frac{8t^2}{2} - 6t + K = 4t^2 - 6t + K$

In which the constant of integration K is the initial velocity, which is v(0). So

$v(t) = 4t^2 - 6t + v(0)$

Position:

$s(t) = \int v(t) dt = \int (4t^2 - 6t + v(0)) dt = \frac{4t^3}{3} - \frac{6t^2}{2} + v(0)t + K = \frac{4t^3}{3} - 3t^2 + v(0)t + K$

The initial position is s(0) = 2. So

$s(t) = \frac{4t^3}{3} - 3t^2 + v(0)t + 2$