Solution 8.4c

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Current revision (16:48, 13 April 2010) (edit) (undo)
 
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Substituting the above values gives a vector equation for <math>t</math>.
Substituting the above values gives a vector equation for <math>t</math>.
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<math>4\textrm{.}8\mathbf{i}-6\textrm{.}8\mathbf{j}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j}+(4\mathbf{i}– 8\mathbf{j})t\ \</math>
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<math>4\textrm{.}8\mathbf{i}-6\textrm{.}8\mathbf{j}=4\mathbf{i}-8\mathbf{j}+(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j})t\ \</math>
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0r <math>4\textrm{.}8\mathbf{i}-6\textrm{.}8\mathbf{j}=(4t+0\textrm{.}2\ )\mathbf{i}– (8t+0\textrm{.}3\ )\mathbf{j}\ \</math>
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0r <math>4\textrm{.}8\mathbf{i}-6\textrm{.}8\mathbf{j}=(4+0\textrm{.}2t\ )\mathbf{i}+ (0\textrm{.}3t-8\ )\mathbf{j}\ \</math>
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Is there a <math>t</math> which satisfies both the <math>\mathbf{i}</math> terms and the <math>\mathbf{j}</math> terms?
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<math>4\textrm{.}8\mathbf{i}=(4+0\textrm{.}2t\ )\mathbf{i}\ \</math>
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and
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<math>-6\textrm{.}8\mathbf{j}=(0\textrm{.}3t-8\ )\mathbf{j}\ \</math>
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It is easily seen that <math>t=4 \ \text{s}</math> satisfies both these equations.

Current revision

We have \displaystyle \mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \ \displaystyle \mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \ and \displaystyle \ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}.

The most suitable equation to use is \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t\ \.

Substituting the above values gives a vector equation for \displaystyle t.

\displaystyle 4\textrm{.}8\mathbf{i}-6\textrm{.}8\mathbf{j}=4\mathbf{i}-8\mathbf{j}+(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j})t\ \

0r \displaystyle 4\textrm{.}8\mathbf{i}-6\textrm{.}8\mathbf{j}=(4+0\textrm{.}2t\ )\mathbf{i}+ (0\textrm{.}3t-8\ )\mathbf{j}\ \

Is there a \displaystyle t which satisfies both the \displaystyle \mathbf{i} terms and the \displaystyle \mathbf{j} terms?

\displaystyle 4\textrm{.}8\mathbf{i}=(4+0\textrm{.}2t\ )\mathbf{i}\ \

and

\displaystyle -6\textrm{.}8\mathbf{j}=(0\textrm{.}3t-8\ )\mathbf{j}\ \

It is easily seen that \displaystyle t=4 \ \text{s} satisfies both these equations.

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