Ian: New page: Using the result from part a) for the acceleration, $\begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( \frac{2t}{5} \right)}dt \\ & \\ & =\frac{{{t}^{2}}}{5}+c \\ & \\ & t=...$

Ian — Mon, 11 Oct 2010 13:48:35 GMT

New page: Using the result from part a) for the acceleration, <math>\begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( \frac{2t}{5} \right)}dt \\ & \\ & =\frac{{{t}^{2}}}{5}+c \\ & \\ & t=...

New page

Using the result from part a) for the acceleration,

<math>\begin{align}
& v=\int{a}dt \\
& \\
& =\int{\left( \frac{2t}{5} \right)}dt \\
& \\
& =\frac{{{t}^{2}}}{5}+c \\
& \\
& t=0,\ v=8\ \Rightarrow \ c=8 \\
& \\
& v=\frac{{{t}^{2}}}{5}+8
\end{align}</math>

Solution 19.4b - Revision history

Ian: New page: Using the result from part a) for the acceleration, \begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( \frac{2t}{5} \right)}dt \\ & \\ & =\frac{{{t}^{2}}}{5}+c \\ & \\ & t=...

Ian: New page: Using the result from part a) for the acceleration, $\begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( \frac{2t}{5} \right)}dt \\ & \\ & =\frac{{{t}^{2}}}{5}+c \\ & \\ & t=...$