In the figure shown, F is in newton and t in seconds. Take $g=10 \mathrm{m}/{\mathrm{s}}^2$. Plot acceleration of the block versus time graph. we are given $g=10 \mathrm{m}/{\mathrm{s}}^2$ and ${\mu }_k$, where ${\mu }_s=0.6$ and ${\mu }_h=0.4$.

$R=mg=20 \mathrm{N}$

$$\begin{gathered} {\mu }_{s}R=0.6\times 20=12 \mathrm{N} \\ {\mu }_{k}R=0.4\times 20=8 \mathrm{N} \end{gathered}$$

Up to 6s, situation is same but after 6s, a constant kinetic friction of 8N will act. 

At 6s, friction will suddenly change from$12 \mathrm{N}\left(={\mu }_{k}R\right)$ to $8 \mathrm{N}\left(={\mu }_{k}R\right)$ and direction of friction is opposite to its motion. Therefore, at 6 s it will start with an initial acceleration.

$a_i=\frac{\text{ decrease in friction }}{\text{ mass }}=\frac{12-8}{2}=2 \mathrm{m}/{\mathrm{s}}^2$

For$t\le 6 \mathrm{s}$

f=F=2 t                                                              …………………(i)

$$\begin{gathered} F_{\text{net}}=F-f=0 \\ a=\frac{F_{\text{net }}}{m}=0 \end{gathered}$$

For $t\ge 68$

F =2 t 

$$\begin{gathered} f={\mu }_{k}R=8 \mathrm{N} \\ {F}_{\text{net }}=F-f=2t-8 \\ \alpha =\frac{F_{\text{met }}}{m}=\frac{2t-8}{2}=(t-4) \end{gathered}$$

At 6s, we can see that, $a_j=2 \mathrm{m}/{\mathrm{s}}^2$

 Further, a-t graph is a straight line of slope =1 and intercept = - 4. Corresponding a-t graph is as shown in figure.