Ian: New page: $\begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( 8-\frac{4t}{5} \right)}dt \\ & \\ & =8t-\frac{2{{t}^{2}}}{5}+c \\ & \\ & t=0,\ v=0\ \Rightarrow \ c=0 \\ & \\ & v=8t-\f...$

Ian — Mon, 11 Oct 2010 09:31:07 GMT

New page: <math>\begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( 8-\frac{4t}{5} \right)}dt \\ & \\ & =8t-\frac{2{{t}^{2}}}{5}+c \\ & \\ & t=0,\ v=0\ \Rightarrow \ c=0 \\ & \\ & v=8t-\f...

New page

<math>\begin{align}
& v=\int{a}dt \\
& \\
& =\int{\left( 8-\frac{4t}{5} \right)}dt \\
& \\
& =8t-\frac{2{{t}^{2}}}{5}+c \\
& \\
& t=0,\ v=0\ \Rightarrow \ c=0 \\
& \\
& v=8t-\frac{2{{t}^{2}}}{5} \\
& \\
& v(10)=8\times 10-\frac{2\times {{10}^{2}}}{5}=40\text{ m}{{\text{s}}^{\text{-1}}}
\end{align}</math>

Solution 19.3b - Revision history

Ian: New page: \begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( 8-\frac{4t}{5} \right)}dt \\ & \\ & =8t-\frac{2{{t}^{2}}}{5}+c \\ & \\ & t=0,\ v=0\ \Rightarrow \ c=0 \\ & \\ & v=8t-\f...

Ian: New page: $\begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( 8-\frac{4t}{5} \right)}dt \\ & \\ & =8t-\frac{2{{t}^{2}}}{5}+c \\ & \\ & t=0,\ v=0\ \Rightarrow \ c=0 \\ & \\ & v=8t-\f...$