Solution 8.5a
From Mechanics
(Difference between revisions)
(New page: We assume the origin is att the point of tacke off which means <math>\ \ \mathbf{ r}_{0}=0</math>. Also we have <math>\mathbf{a}=\mathbf{i}+4\mathbf{j},\ \</math> <math>\mathbf{u}=80\ma...) |
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<math>\mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}</math> | <math>\mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}</math> | ||
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| + | Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> gives at time <math>t</math>, | ||
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| + | <math>\mathbf{r}=80\mathbf{i}t+\frac{1}{2}(\mathbf{i}+4\mathbf{j}){t}^{\ 2}</math> | ||
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| + | <math>\mathbf{r}=(80t+\frac{1}{2}{t}^{\ 2}\ )\mathbf{i}+2{t}^{\ 2}\ \mathbf{j}</math> | ||
Current revision
We assume the origin is att the point of tacke off which means \displaystyle \ \ \mathbf{ r}_{0}=0.
Also we have \displaystyle \mathbf{a}=\mathbf{i}+4\mathbf{j},\ \ \displaystyle \mathbf{u}=80\mathbf{i}\ \.
Using \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t\ \ gives at time \displaystyle t,
\displaystyle \mathbf{v}=80\mathbf{i}+(\mathbf{i}+4\mathbf{j})t\ \ or
\displaystyle \mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}
Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} gives at time \displaystyle t,
\displaystyle \mathbf{r}=80\mathbf{i}t+\frac{1}{2}(\mathbf{i}+4\mathbf{j}){t}^{\ 2}
\displaystyle \mathbf{r}=(80t+\frac{1}{2}{t}^{\ 2}\ )\mathbf{i}+2{t}^{\ 2}\ \mathbf{j}
