Answer:
[tex]s(t) = \frac{4t^3}{3} - 3t^2 + v(0)t + 2[/tex]
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:
[tex]a(t) = 8t - 6[/tex]
Velocity:
[tex]v(t) = \int a(t) dt = \int (8t - 6) dt = \frac{8t^2}{2} - 6t + K = 4t^2 - 6t + K[/tex]
In which the constant of integration K is the initial velocity, which is v(0). So
[tex]v(t) = 4t^2 - 6t + v(0)[/tex]
Position:
[tex]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[/tex]
The initial position is s(0) = 2. So
[tex]s(t) = \frac{4t^3}{3} - 3t^2 + v(0)t + 2[/tex]
