# What is the position of a particle at time t=2 if a particle moves along the x axis so that at any time t>0, the velocity is given by v(t)=4-6t^2, and at position x=7 at t=1?

Feb 5, 2015

The answer is: $x = - 3$.

The law that gives the position of the particle at the time $t$ is given by the integral:

$x \left(t\right) = \int \left(4 - 6 {t}^{2}\right) \mathrm{dt} = 4 t - 6 {t}^{3} / 3 + c = 4 t - 2 {t}^{3} + c$.

To find $c$, we can use the intial conditions:

$x \left(1\right) = 7$,

so:

$7 = 4 - 2 + c \Rightarrow c = 5$.

So the law becomes:

$x \left(t\right) = 4 t - 2 {t}^{3} + 5$.

Now we can use it to find the positition at the time $t = 2$:

$x \left(2\right) = 8 - 16 + 5 = - 3$.