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Solution 8.6b

From Mechanics

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(New page: Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math> According to the text <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> We assume th...)
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<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>
<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>
 +
and part a) gave
 +
 +
<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}</math>
We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.
We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.
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At <math>t=10+40=50</math> we obtain
At <math>t=10+40=50</math> we obtain
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+
<math>\mathbf{r}=(\mathbf{i}+2\mathbf{j}) \times 50+\frac{1}{2}(\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j} )\times{{50}^{\ 2}}</math>
The distance is the magnitude of this vector
The distance is the magnitude of this vector

Revision as of 19:35, 14 April 2010

Here we use

r=ut+21at 2+r0

According to the text

u=i+2j

and part a) gave

a=0.5i−j

We assume the starting point is the origin so that r0=0.

At t=10+40=50 we obtain

r=(i+2j)50+21(a=0.5i−j)50 2


The distance is the magnitude of this vector

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