Looking for classes? Ksquare Career Institute, Bengaluru →
The position of particle is given by $\vec{r} = 2t^{2}\hat{i} + 3\hat{j} + 4\hat{k}$, where t is in second and the coefficients have proper units for $\vec{r}$ to be in metre. The $\vec{a}(t)$ of the particle at $t = 1 s$ is
Explanation
$\vec{r} = 2t^{2}\hat{i} + 3\hat{j} + 4\hat{k}$ $\therefore \vec{v} = d\vec{r}/dt = d/dt(2t^{2}\hat{i} + 3t\hat{j} + 4\hat{k}) = 4t\hat{i} + 3\hat{j}$ $\vec{a} = d\vec{v}/dt = d/dt(4t\hat{i} + 3\hat{j}) = 4\hat{i} \therefore \vec{a} = 4 m^{-2}$ along x-direction