The particle has acceleration vector

We're told that it starts off at the origin, so that its position vector at
is

and that it has an initial velocity of 12 m/s in the positive
direction, or equivalently its initial velocity vector is

To find the velocity vector for the particle at time
, we integrate the acceleration vector:

![\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cleft%5B12%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cjmath)
![\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cleft%5B12%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5C%2C%5Cvec%5Cjmath)
Then we integrate this to find the position vector at time
:

![\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%2812%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cjmath)
![\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cleft%5B%5Cleft%2812%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29t%2B%5Cleft%28-1.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%282.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5C%2C%5Cvec%5Cjmath)
Solve for the time when the
coordinate is 18 m:

At this point, the
coordinate is

so the answer is C.