The graph below shows the variation of a force F with time t on a...

Given force varies as

$F ∝ {t}^{n} ⇒ F = k{t}^{n} . . . . \left(i\right)$

From graph, we observe

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

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

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

$16 = k\left(4\right){}^{n} and$2 = k\left(2\right){}^{n}$</p> <p>$⇒ 8 = \dfrac{{2}^{2n}}{{2}^{n}} ⇒ n = 3$</p> <p>Also,$2 = k{\left(2\right)}^{n} ⇒ 2 = k\left(2\right){}^{3} ⇒ k = \dfrac{1}{4}$</p> <p>So, force is given by$F = \dfrac{{t}^{3}}{4}$</p> <p>Now,$F = \dfrac{dp}{dt},$(where,$p =$momentum)</p> <p>$⇒ dp = Fdt ⇒ {∫}_{{v}_{i}}^{{v}_{f}}dp = {∫}_{{t}_{i}}^{{t}_{f}}Fdt$</p> <p>$⇒ m\left({v}_{f} - {v}_{i}\right) = {∫}_{{t}_{i}}^{{t}_{f}}\dfrac{{t}^{3}}{4} \cdot dt$</p> <p>$⇒ m\left({v}_{f} - {v}_{i}\right) = \dfrac{{t}_{f}^{4} - {t}_{i}^{4}}{16} . . . . \left(ii\right)$</p> <p>Now, when${t}_{i} = 0 and ${t}_{f} = 3s$
We have, ${v}_{i} = 0$ and ${v}_{f} = 2{ms}^{- 1}$
So, from Eq. (ii), we have

$2m = \dfrac{81}{16} . . . . \left(iii\right)$

and when we take ${t}_{i} = 0, {t}_{f} = 4s, {v}_{i} = 0$

and ${v}_{f} = v$ (let), then again from Eq (ii) we gives

$mv = \dfrac{256}{16} . . . . \left(iv\right)$

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

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

$⇒ v = \dfrac{256 \times 2}{81} \approx 6 . 5{ms}^{- 1}$