Ian: New page: The expression for the acceleration $\mathbf{a}$ has been obtained in part a), $\begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( \frac{3{{t}^{2}}}{2}+{{c...$

Ian — Mon, 11 Oct 2010 16:28:09 GMT

New page: The expression for the acceleration <math>\mathbf{a}</math> has been obtained in part a), <math>\begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( \frac{3{{t}^{2}}}{2}+{{c...

New page

The expression for the acceleration <math>\mathbf{a}</math> has been obtained in part a),

<math>\begin{align}
& \mathbf{v}=\int{\mathbf{a}dt} \\
& \\
& =\left( \frac{3{{t}^{2}}}{2}+{{c}_{1}} \right)\mathbf{i}+\left( \frac{3t}{2}-\frac{5{{t}^{2}}}{8}+{{c}_{2}} \right)\mathbf{j} \\
& \\
& t=0,\ \mathbf{v}=-8\mathbf{i}\ \Rightarrow \ {{c}_{1}}=-8,\ {{c}_{2}}=0 \\
& \\
& \mathbf{v}=\left( \frac{3{{t}^{2}}}{2}-8 \right)\mathbf{i}+\left( \frac{3t}{2}-\frac{5{{t}^{2}}}{8} \right)\mathbf{j}
\end{align}</math>

(Note in the above when integrating a vector a constant of integration has to be introduced for each component).

Solution 19.7b - Revision history

Ian: New page: The expression for the acceleration \mathbf{a} has been obtained in part a), \begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( \frac{3{{t}^{2}}}{2}+{{c...

Ian: New page: The expression for the acceleration $\mathbf{a}$ has been obtained in part a), $\begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( \frac{3{{t}^{2}}}{2}+{{c...$