Find the velocity vector and position vector of a particle if its acceleration is given by $\mathbf{a}( t)\,{=}$ $-9.8\kern.7pt\mathbf{j}{m}/\!{s}^{2}$, $\mathbf{ v}( 0)\;=\;9.8\kern.7pt\mathbf{j}{m}/\!{s}$, and $\mathbf{r}( 0)\;=\;\mathbf{0} {m}$ for $0\leq t\leq 2.$

Solution The velocity vector $\mathbf{v}( t)$ is $$\begin{equation*} \mathbf{v}( t)\;=\;\int \mathbf{a}( t) dt=\left( \int -9.8dt\right) \mathbf{j}=c_{1}\mathbf{i}+\left( -9.8t+c_{2}\right) \mathbf{j} \end{equation*}$$

Since $\mathbf{v}( 0)\;=\;9.8\mathbf{j},$ then $c_{1}=0$ and $c_{2}=9.8,$ and the velocity vector is $$\begin{equation*} \mathbf{v}( t)\;=\;\left( -9.8t+9.8\right) \mathbf{j} \end{equation*}$$

The position vector of the particle is given by $$\begin{equation*} \mathbf{r}( t)\;=\;\int \mathbf{v}( t) dt=\left[ \int \left( -9.8t+9.8\right) dt\right] \mathbf{j}=d_{1}\mathbf{i}+\left( -4.9t^{2}+9.8t+d_{2}\right) \mathbf{j} \end{equation*}$$

Since $\mathbf{r}( 0)\;=\;\mathbf{0},$ then $d_{1}=0$ and $d_{2}=0,$ and the position of the particle is given by $$\begin{equation*} \mathbf{r}( t)\;=\;( -4.9t^{2}+9.8t) \mathbf{j} \notag \end{equation*}$$