Solution 19.3b
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(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...)
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\displaystyle \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}
