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Solution 8.1a

From Mechanics

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(New page: In this case we have <math>\mathbf{u}=4\mathbf{i}</math> , <math>\mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j}</math> and <math>{{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}</math>. Substit...)
Line 7: Line 7:
Substituting these into the equation
Substituting these into the equation
-
<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>
+
<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+400\mathbf{i}+350\mathbf{j}</math>
gives the position vector of the ball at time <math>10</math> as:
gives the position vector of the ball at time <math>10</math> as:
<math>\begin{align}
<math>\begin{align}
-
& \mathbf{r}=(4\mathbf{i})t+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j}){{t}^{\ 2}}+1\textrm{.}5\mathbf{j} \\
+
& \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\
& =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\
& =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\
& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}
& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}
\end{align}</math>
\end{align}</math>

Revision as of 17:15, 12 April 2010

In this case we have u=4i , a=09i+07j and r0=400i+350j.

Substituting these into the equation r=ut+21at 2+400i+350j gives the position vector of the ball at time 10 as:

r=(4i)10+21(09i+07j)10 2+400i+350j=8ti+2tj5t 2j+1.5j=(8t)i+(1.5+2t5t 2)j m