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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.

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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,

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