1. Use impulse-momentum theorem

Since the particle starts from rest and the force has constant direction, the final velocity after time $2T$ is obtained from

$M v = \int_0^{2T} F(t)\,dt$

Given

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

So,

$v=\frac{1}{M}\int_0^{2T} F_0\left[1-\left(\frac{t-T}{T}\right)^2\right]dt$

$v=\frac{F_0}{M}\int_0^{2T}\left[1-\frac{(t-T)^2}{T^2}\right]dt$

2. Evaluate the integral

Let

$x=t-T \Rightarrow dx=dt$

When $t=0$, $x=-T$; when $t=2T$, $x=T$.

Thus,

$v=\frac{F_0}{M}\int_{-T}^{T}\left(1-\frac{x^2}{T^2}\right)dx$

Now,

$\int_{-T}^{T}1\,dx=2T$

and

$\int_{-T}^{T}\frac{x^2}{T^2}dx=\frac{1}{T^2}\int_{-T}^{T}x^2dx$

Since $x^2$ is even,

$\int_{-T}^{T}x^2dx=2\int_0^T x^2dx=2\cdot\frac{T^3}{3}=\frac{2T^3}{3}$

Hence,

$\int_{-T}^{T}\frac{x^2}{T^2}dx=\frac{1}{T^2}\cdot \frac{2T^3}{3}=\frac{2T}{3}$

Therefore,

$\int_{-T}^{T}\left(1-\frac{x^2}{T^2}\right)dx=2T-\frac{2T}{3}=\frac{4T}{3}$

3. Find the final velocity

$v=\frac{F_0}{M}\cdot \frac{4T}{3}$

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

4. Match with options

This corresponds to:

$\boxed{\text{Option C}}$

5. Compare with stored correct answer

Stored correct answer: C

Our derived answer is also C, so they agree.