Paul's Online Notes
Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / Velocity and Acceleration
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 1-11 : 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.

Show All Steps Hide All Steps

Start Solution

To 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 2

The 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,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left( {\frac{1}{2}{t^3} - 5} \right)\,\vec i + \left( { - 4{{\bf{e}}^{ - t}} - 3t + 6} \right)\,\vec j + \left( {{t^4} - 3t - 3} \right)\vec k}}\]