Section 12.11 : Velocity and Acceleration
1. An objects acceleration is given by \(\vec a = 3t\,\vec i - 4{{\bf{e}}^{ - t}}\,\vec j + 12{t^2}\vec k\). The objects initial velocity is \(\vec v\left( 0 \right) = \vec j - 3\vec k\) and the objects initial position is \(\vec r\left( 0 \right) = - 5\vec i + 2\vec j - 3\vec k\). Determine the objects velocity and position functions.
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Start SolutionTo determine the velocity function all we need to do is integrate the acceleration function.
\[\vec v\left( t \right) = \int{{3t\,\vec i - 4{{\bf{e}}^{ - t}}\,\vec j + 12{t^2}\vec k\,dt}} = \frac{3}{2}{t^2}\,\vec i + 4{{\bf{e}}^{ - t}}\,\vec j + 4{t^3}\vec k + \vec c\]Don’t forget the “constant” of integration, which in this case is actually the vector \(\vec c = {c_1}\vec i + {c_2}\vec j + {c_3}\vec k\) . To determine the constant of integration all we need is to use the value \(\vec v\left( 0 \right)\) that we were given in the problem statement.
\[\vec j - 3\vec k = \vec v\left( 0 \right) = 4\,\vec j + {c_1}\vec i + {c_2}\vec j + {c_3}\vec k\,\]To determine the values of \({c_1}\), \({c_2}\), and \({c_3}\) all we need to do is set the various components equal.
\[\begin{align*}&{\vec i:0 = {c_1}}\\&{\vec j:1 = 4 + {c_2}}\\&{\vec k: - 3 = {c_3}}\end{align*}\hspace{0.5in} \Rightarrow \hspace{0.5in}{c_1} = 0,\,\,\,\,{c_2} = - 3,\,\,\,\,{c_3} = - 3\]The velocity is then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec v\left( t \right) = \frac{3}{2}{t^2}\,\vec i + \left( {4{{\bf{e}}^{ - t}} - 3} \right)\,\vec j + \left( {4{t^3} - 3} \right)\vec k}}\] Show Step 2The position function is simply the integral of the velocity function we found in the previous step.
\[\vec r\left( t \right) = \int{{\frac{3}{2}{t^2}\,\vec i + \left( {4{{\bf{e}}^{ - t}} - 3} \right)\,\vec j + \left( {4{t^3} - 3} \right)\vec k\,dt}} = \frac{1}{2}{t^3}\,\vec i + \left( { - 4{{\bf{e}}^{ - t}} - 3t} \right)\,\vec j + \left( {{t^4} - 3t} \right)\vec k + \vec c\] We’ll use the value of \(\vec r\left( 0 \right)\) from the problem statement to determine the value of the constant of integration. \[ - 5\vec i + 2\vec j - 3\vec k = \vec r\left( 0 \right) = - 4\,\vec j + {c_1}\vec i + {c_2}\vec j + {c_3}\vec k\] \[\begin{align*} &{\vec i: - 5 = {c_1}}\\& {\vec j:2 = - 4 + {c_2}}\\ &{\vec k: - 3 = {c_3}}\end{align*}\hspace{0.5in} \Rightarrow \hspace{0.25in}{c_1} = - 5,\,\,\,\,{c_2} = 6,\,\,\,\,{c_3} = - 3\]The position function is then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left( {\frac{1}{2}{t^3} - 5} \right)\,\vec i + \left( { - 4{{\bf{e}}^{ - t}} - 3t + 6} \right)\,\vec j + \left( {{t^4} - 3t - 3} \right)\vec k}}\]