$(b)$ Given force varies as

$F \propto t^{n} \Rightarrow F=k t^{n} \quad \dots(i)$

From graph, we observe

at $t=2 s , F=2 N$

and at $t=4 s , F=16 N$

So, putting the values in Eq $(i)$, we have

$16=k(4)^{n} \text { and } 2=k(2)^{n}$

$\Rightarrow \quad 8=\frac{2^{2 n}}{2^{n}} \Rightarrow n=3$

$\text { Also, } 2=k(2)^{n} \Rightarrow 2=k(2)^{3} \Rightarrow k=\frac{1}{4}$

So, force is given by $F=\frac{t^{3}}{4}$

Now, $F=\frac{d p}{d t},($ where, $p=$ momentum $)$

$\Rightarrow \quad d p=F d t \Rightarrow \int \limits_{v_{i}}^{v_{f}} d p=\int \limits_{\varepsilon_{i}}^{t_{f}} F d t$

$\Rightarrow \quad m\left(v_{f}-v_{i}\right)=\int \limits_{t_{i}}^{t_{f}} \frac{t^{3}}{4} \cdot d t$

$\Rightarrow \quad m\left(v_{f}-v_{i}\right)=\frac{t_{f}^{4}-t_{i}^{4}}{16}\quad \dots(ii)$

Now, when $t_{i}=0$ and $t_{f}=3 s$

We have, $v_{i}=0$ and $v_{f}=2 ms ^{-1}$

So, from Eq. (ii), we have

$2 m=\frac{81}{16} \quad \dots(iii)$

and when we take $t_{i}=0, t_{f}=4 s , v_{i}=0$ and $v_{f}=v$ (let), then again from Eq (ii), we gives

$m v=\frac{256}{16} \quad \dots(iv)$

Dividing Eq. $(iv)$ by Eq. $(iii)$, we get

$\frac{v}{2} =\frac{256}{81}$

$\Rightarrow \quad v= \frac{256 \times 2}{81} \approx 6.5 \,ms ^{-1}$