Solution 19.7a
From Mechanics
(Difference between revisions)
(New page: <math>\mathbf{a}=\frac{\mathbf{F}}{4}=\frac{12t\mathbf{i}+(6-5t)\mathbf{j}}{4}=3t\mathbf{i}+\left( \frac{3}{2}-\frac{5t}{4} \right)\mathbf{j}</math>) |
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| - | <math>\mathbf{a}=\frac{\mathbf{F}}{4}=\frac{12t\mathbf{i}+(6-5t)\mathbf{j}}{4}=3t\mathbf{i}+\left( \frac{3}{2}-\frac{5t}{4} \right)\mathbf{j}</math> | + | <math>\mathbf{a}=\frac{\mathbf{F}}{4}= |
| + | \frac{\smash{ {12t\mathbf{i}+(6-5t)\mathbf{j}} }}{4}= | ||
| + | 3t\mathbf{i}+\left( \frac{3}{2}-\frac{5t}{4} \right)\mathbf{j}</math> | ||
Current revision
\displaystyle \mathbf{a}=\frac{\mathbf{F}}{4}= \frac{\smash{ {12t\mathbf{i}+(6-5t)\mathbf{j}} }}{4}= 3t\mathbf{i}+\left( \frac{3}{2}-\frac{5t}{4} \right)\mathbf{j}
