a particle moces along the x-axis with velocity v(t)= 3t^2 +6t. The particle is at position x=2 at t=0, what is the position of the particle at t=0???

Question

mathteacher1729 · Answer

HINT: velocity is the DERIVATIVE of position (with respect to time).

anonymous · Answer

The units do not make sense.

anonymous · Answer

ok so i take the antiderivative then what?

mathteacher1729 · Answer

What do you get when you take the antiderivative?

anonymous · Answer

t^3+3t^2

anonymous · Answer

t^3 + 3t^2

mathteacher1729 · Answer

$$\huge x(t)= t^3 + 3t^2+c$$
Don't forget the constant. ;-p 

Now, they give us that $$\huge x(0)=2$$

So substitute that into the equation x(t) and solve for C.

anonymous · Answer

oh ok so then i get c to equal -20 then i plug 1 in and get the position to be -16?

mathteacher1729 · Answer

Woah, how'd you get -20??