Math Help 2!!!!?

Question

3. A particles moves along the x –axis so that, at any times t>0 its acceleration is given by a(t)=6t +6, At time t=0, the velocity of the particles is -9, and its position is -27.

- Find v(t), the velocity of the particle at any times t>0
- For what values of t>0 is the particles moving to the right?
- Find x(t), the position of the particle at any time t>0.

Answer 1

Integrate the expression for a(t) to get an expression for v(t). use the condition you were given to evaluate the constant of integration

a(t) = 6t + 6

v(t) = 3t^2 + 6t + c

condition
v(t=0) = c = -9

hence
v(t) = 3t^2 + 6t - 9

When the particle is moving to the right, the velocity is positive. Find the values of t that make this expression positive.

Integrate the expression again to get x(t)

x(t) = t^3 + 3t^2 -9t + c

x(t=0) = c = -27

hence
x(t) = t^3 + 3t^2 -9t -27

Answer 2

64