Solution 19.8a
From Mechanics
(Difference between revisions)
(New page: <math>\begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( {{c}_{1}} \right)\mathbf{i}+\left( -2t+{{c}_{2}} \right)\mathbf{j} \\ & \\ & t=0,\ \mathbf{v}=40\mathbf{i}\ \Rig...) |
(New page: <math>\begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( {{c}_{1}} \right)\mathbf{i}+\left( -2t+{{c}_{2}} \right)\mathbf{j} \\ & \\ & t=0,\ \mathbf{v}=40\mathbf{i}\ \Rig...) |
Current revision
\displaystyle \begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( {{c}_{1}} \right)\mathbf{i}+\left( -2t+{{c}_{2}} \right)\mathbf{j} \\ & \\ & t=0,\ \mathbf{v}=40\mathbf{i}\ \Rightarrow \ {{c}_{1}}=40,\ {{c}_{2}}=0 \\ & \\ & \mathbf{v}=\left( 40 \right)\mathbf{i}+\left( -2t \right)\mathbf{j} \end{align}
