Processing Math: Done
Solution 19.5b
From Mechanics
(Difference between revisions)
(New page: Using the expression for <math>v</math> obtained in part a), <math>\begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{12...) |
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| Line 2: | Line 2: | ||
<math>\begin{align} | <math>\begin{align} | ||
| - | & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{ | + | & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{320} \right)}dt \\ |
& \\ | & \\ | ||
| - | & =\left[ 20t-\frac{{{t}^{3}}}{ | + | & =\left[ 20t-\frac{{{t}^{3}}}{960} \right]_{0}^{40} \\ |
& \\ | & \\ | ||
| - | & =\left( 20\times 40-\frac{{{40}^{3}}}{ | + | & =\left( 20\times 40-\frac{{{40}^{3}}}{960} \right)-0 \\ |
& \\ | & \\ | ||
| - | & = | + | & =733\text{ m} |
\end{align}</math> | \end{align}</math> | ||
Current revision
Using the expression for
040
20−3t2320
dt=
20t−t3960
040=20
40−403960−0=733 m
