The position of particle is given by $$\vec r = 2{t^2}\hat i + 3t\hat j + 4\hat k,$$     where $$t$$ is in second and the coefficients have proper units for $${\vec r}$$ to be in metre. The $$\vec a\left( t \right)$$  of the particle at $$t = 1\,s$$   is

Solution :
$$\eqalign{ & \vec r = 2{t^2}\hat i + 3t\hat j + 4\hat k \cr & \therefore \vec v = \frac{{d\vec r}}{{dt}} = \frac{d}{{dt}} = \left( {2{t^2}\hat i + 3t\hat j + 4\hat k} \right) = 4t\hat i + 3\hat j \cr & \vec a = \frac{{d\vec v}}{{dt}} = \frac{d}{{dt}}\left( {4t\hat i + 3\hat j} \right) = 4\hat i \cr} $$
$$\therefore \vec a = 4\,m{s^{ - 2}}$$   along $$x$$-direction