Solution 19.1c

From Mechanics

(Difference between revisions)
Jump to: navigation, search
(New page: <math>\begin{align} & s(15)=\int_{0}^{15}{v}dt \\ & \\ & =\int_{0}^{15}{( t-\frac{{{t}^{2}}}{30} )dt} \\ & \\ & =\left[ \frac{{{t}^{2}}}{2}-\frac{{{t}^{3}}}{90} \right]_{0}^{15} \\ & ...)
Current revision (09:35, 11 October 2010) (edit) (undo)
 
Line 1: Line 1:
 +
Using the result for the velocity obtained in part a)
 +
<math>\begin{align}
<math>\begin{align}
& s(15)=\int_{0}^{15}{v}dt \\
& s(15)=\int_{0}^{15}{v}dt \\

Current revision

Using the result for the velocity obtained in part a)

\displaystyle \begin{align} & s(15)=\int_{0}^{15}{v}dt \\ & \\ & =\int_{0}^{15}{( t-\frac{{{t}^{2}}}{30} )dt} \\ & \\ & =\left[ \frac{{{t}^{2}}}{2}-\frac{{{t}^{3}}}{90} \right]_{0}^{15} \\ & \\ & =\left( \frac{{{15}^{2}}}{2}-\frac{{{15}^{3}}}{90} \right)-0 \\ & \\ & =75\text{ m} \end{align}