A particle moves along a straight line and its velocity depends on time $t$ as $\nu =4t-t^2,$ where time $t$ is measured in seconds and distance in metres. Then for first 5 seconds

$v=0\Rightarrow t=0,4\text{s}$

In first four seconds, $v>0$ .
Displacement in first 4 seconds,
 $d_1=\underset{0}{\overset{4}{\int }}vdt=\underset{0}{\overset{4}{\int }}\left(4t-t^2\right)dt={\left.\left(2t^2-\frac{t^3}{3}\right)\right|}_0^4=\left(32-\frac{64}{3}\right)=\frac{32}{3}m$
In the next one second, $v<0$  
Displacement from $t=4\text{s}$  to  $t=5\text{s}$

$d_2=\underset{4}{\overset{5}{\int }}vdt=\underset{4}{\overset{5}{\int }}\left(4t-t^2\right)dt={\left.\left(2t^2-\frac{t^3}{3}\right)\right|}_4^5=\left(50-\frac{125}{3}\right)-\left(32-\frac{64}{3}\right)=\frac{25}{3}-\frac{32}{3}$ 
$d_2=\frac{-7}{3}m$

Average speed $=\frac{\text{Total\hspace{0.17em}\hspace{0.17em}distance}}{\text{time}}=\frac{d_1+\left|d_2\right|}{t}=\frac{\left(\frac{32}{3}\right)+\left(\frac{7}{3}\right)}{5}=\frac{39}{3\times 5}=\frac{13}{5}m{s}^{-1}$
Average velocity $=\frac{\text{Total\hspace{0.17em}\hspace{0.17em}diplacement}}{\text{time}}=\frac{d_1+d_2}{t}=\frac{\left(\frac{32}{3}\right)-\left(\frac{7}{3}\right)}{5}=\frac{25}{3\times 5}=\frac{5}{3}m{s}^{-1}$

$a=4-2t$
At  $t=0,a=4m/s^2$