Processing Math: Done
Solution to Test Paper 1
From Mechanics
(Difference between revisions)
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- | | 5 (a) | + | | 5 (a) |
- | | | + | | <math>\begin{align} |
- | | | + | & 5\mathbf{i}-2\mathbf{j}=4\mathbf{i}+3\mathbf{j}+10\mathbf{a} \\ |
+ | & \mathbf{a}=\frac{1}{10}(\mathbf{i}-5\mathbf{j})=\left( 0\textrm{.}1\mathbf{i}-0\textrm{.}5\mathbf{j} \right)\text{ m}{{\text{s}}^{\text{-2}}} \\ | ||
+ | \end{align}</math> | ||
+ | | (3 marks) | ||
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| 5 (b) | | 5 (b) | ||
- | | | + | | <math>\mathbf{r}=(4\mathbf{i}+3\mathbf{j})t+0\textrm{.}5(0\textrm{.}1\mathbf{i}-0\textrm{.}5\mathbf{j}){{t}^{2}}</math> |
- | | | + | | (2 marks) |
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| 5 (c) | | 5 (c) | ||
- | | | + | | <math>\begin{align} |
- | | | + | & \mathbf{r}=(4t+0\textrm{.}05{{t}^{2}})\mathbf{i}+(3t-0\textrm{.}25{{t}^{2}})\mathbf{j} \\ |
+ | & 3t-0\textrm{.}25{{t}^{2}}=0 \\ | ||
+ | & t(3-0\textrm{.}25t)=0 \\ | ||
+ | & t=0\text{ or }t=\frac{3}{0\textrm{.}25}=12\text{ s} \\ | ||
+ | & t=12\text{ s} \\ | ||
+ | \end{align}</math> | ||
+ | | (3 marks) | ||
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| 5 (d) | | 5 (d) | ||
- | | | + | | <math>\begin{align} |
- | | | + | & \mathbf{v}=(4+0\textrm{.}1t)\mathbf{i}+(3-0\textrm{.}5t)\mathbf{j} \\ |
+ | & 3-0\textrm{.}5t=0 \\ | ||
+ | & t=6 \\ | ||
+ | \end{align}</math> | ||
+ | | (3 marks) | ||
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- | | | + | | '''(11 marks)''' |
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