Question

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???

Answer 1

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

Answer 2

The units do not make sense.

Answer 3

ok so i take the antiderivative then what?

Answer 4

What do you get when you take the antiderivative?

Answer 5

t^3+3t^2

Answer 6

t^3 + 3t^2

Answer 7

\[\huge x(t)= t^3 + 3t^2+c\]
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.

Answer 8

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

Answer 9

Woah, how'd you get -20??