Solution 19.7b
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(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...)
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The expression for the acceleration \displaystyle \mathbf{a} has been obtained in part a),
\displaystyle \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}
(Note in the above when integrating a vector a constant of integration has to be introduced for each component).
