Finding the Velocity and Position of a Particle

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*}