A particle moves along a straight line with an acceleration given by a(t) = 6t - 4. If its initial velocity is v(0) = 10 and its initial position is s(0) = 0, find the position function s(t). Explain your calculations.

To find the position function s(t), we need to integrate the acceleration function a(t) twice, since acceleration is the second derivative of position with respect to time.

First, let's find the velocity function v(t). We know that acceleration is the first derivative of velocity, so we can write:

v(t) = ∫a(t) dt

Integrating the given acceleration function (6t - 4) with respect to t, we get:

v(t) = ∫(6t - 4) dt = 3t^2 - 4t + C1

Now, we use the initial condition v(0) = 10 to solve for the constant C1:

10 = 3(0)^2 - 4(0) + C1
C1 = 10

So our velocity function is:

v(t) = 3t^2 - 4t + 10

Next, we need to find the position function s(t). We know that velocity is the first derivative of position, so we can write:

s(t) = ∫v(t) dt

Integrating the velocity function (3t^2 - 4t + 10) with respect to t, we get:

s(t) = ∫(3t^2 - 4t + 10) dt = t^3 - 2t^2 + 10t + C2

Now, we use the initial condition s(0) = 0 to solve for the constant C2:

0 = (0)^3 - 2(0)^2 + 10(0) + C2
C2 = 0

So our position function is:

s(t) = t^3 - 2t^2 + 10t

In summary, we found the position function s(t) by integrating the acceleration function a(t) twice and using the given initial conditions for velocity and position. The final position function is s(t) = t^3 - 2t^2 + 10t.