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(D) Given that the particle is initially at rest,so at $t = 0$,$u = 0$.According to Newton's second law,$F = M a$,so $a = \frac{F}{M}$.Substituting th

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$\frac{4 F_{0} T}{3 M}$

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(D) Given that the particle is initially at rest,so at $t = 0$,$u = 0$.According to Newton's second law,$F = M a$,so $a = \frac{F}{M}$.Substituting the given force relation: $a = \frac{F_{0}}{M} \left(1 - \frac{(t - T)^{2}}{T^{2}}\right) = \frac{dv}{dt}$.To find the velocity $v$ at $t = 2T$,we integrate the acceleration with respect to time from $t = 0$ to $t = 2T$:$v = \int_{0}^{2T} \frac{F_{0}}{M} \left(1 - \frac{(t - T)^{2}}{T^{2}}\right) dt$.Let $x = t - T$,then $dx = dt$. When $t = 0, x = -T$ and when $t = 2T, x = T$.$v = \frac{F_{0}}{M} \int_{-T}^{T} \left(1 - \frac{x^{2}}{T^{2}}\right) dx$.$v = \frac{F_{0}}{M} \left[ x - \frac{x^{3}}{3T^{2}} \right]_{-T}^{T}$.$v = \frac{F_{0}}{M} \left( (T - \frac{T^{3}}{3T^{2}}) - (-T - \frac{(-T)^{3}}{3T^{2}}) \right)$.$v = \frac{F_{0}}{M} \left( (T - \frac{T}{3}) - (-T + \frac{T}{3}) \right) = \frac{F_{0}}{M} \left( \frac{2T}{3} + \frac{2T}{3} \right) = \frac{4 F_{0} T}{3 M}$.

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