A particle of mass $$M$$ originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation $$F = F_0\left[1 - \left(\frac{t-T}{T}\right)^2\right]$$ where $$F_0$$ and $$T$$ are constants. The force acts only for the time interval $$2T$$. The velocity $$v$$ of the particle after time $$2T$$ is:

We have a particle of mass $$M$$ that is initially at rest, so its initial velocity is zero. A time-dependent force acts on it along a fixed direction for the duration $$0 \le t \le 2T$$. The magnitude of the force is given by

$$F(t)=F_0\left[1-\left(\frac{t-T}{T}\right)^2\right].$$

According to Newton’s second law, the relation between force and acceleration is

$$F(t)=M\,a(t)=M\,\frac{dv}{dt}.$$

We can rewrite this as

$$\frac{dv}{dt}=\frac{F(t)}{M}.$$

To obtain the velocity after the force has acted for the full interval, we integrate both sides with respect to time from the initial instant $$t=0$$ to the final instant $$t=2T$$. Because the particle starts from rest, its initial velocity is zero. Hence

$$v(2T)-v(0)=\int_{0}^{2T}\frac{dv}{dt}\,dt=\int_{0}^{2T}\frac{F(t)}{M}\,dt.$$

Since $$v(0)=0$$, the velocity after time $$2T$$ is simply

$$v=\frac{1}{M}\int_{0}^{2T}F(t)\,dt.$$

Substituting the given expression for $$F(t)$$, we have

$$v=\frac{1}{M}\int_{0}^{2T}F_0\left[1-\left(\frac{t-T}{T}\right)^2\right]dt.$$

Now we factor out the constant $$F_0$$:

$$v=\frac{F_0}{M}\int_{0}^{2T}\left[1-\frac{(t-T)^2}{T^2}\right]dt.$$

To simplify the integral, let us change variables. Define

$$x=t-T \quad\Longrightarrow\quad t=x+T.$$

When $$t=0$$, we get $$x=-T$$; and when $$t=2T$$, we get $$x=+T$$. In terms of the new variable, the integral becomes

$$v=\frac{F_0}{M}\int_{-T}^{T}\left[1-\frac{x^2}{T^2}\right]dx.$$

We now carry out the integration term by term:

$$\int_{-T}^{T}1\,dx = \Bigl[x\Bigr]_{-T}^{T}=T-(-T)=2T,$$

$$\int_{-T}^{T}\frac{x^2}{T^2}\,dx = \frac{1}{T^2}\int_{-T}^{T}x^2\,dx = \frac{1}{T^2}\left[\frac{x^3}{3}\right]_{-T}^{T} = \frac{1}{T^2}\left(\frac{T^3}{3}-\frac{(-T)^3}{3}\right)=\frac{1}{T^2}\left(\frac{T^3}{3}+\frac{T^3}{3}\right)=\frac{2T}{3}.$$

Hence the complete integral evaluates to

$$\int_{-T}^{T}\left[1-\frac{x^2}{T^2}\right]dx \;=\; 2T-\frac{2T}{3}=\frac{4T}{3}.$$

Substituting this result back, we get

$$v=\frac{F_0}{M}\left(\frac{4T}{3}\right)=\frac{4F_0T}{3M}.$$

Therefore the speed of the particle after the force has acted for the time interval $$2T$$ is

$$v=\frac{4F_0T}{3M}.$$

Hence, the correct answer is Option C.