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)[1-((t-T)/(T))^(2)]`
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 :

Options

A. `2F_(0)T//"M"`
B. `F_(0)T//"2M"`
C. `4F_(0)T//"3M"`
D. `F_(0)T//"3M"`

Answer

To solve the problem, we need to find the velocity \( v \) of a particle of mass \( M \) after a time interval of \( 2T \) when subjected to a time-varying force given by:

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

### Step-by-Step Solution:

1. **Understanding the Force**:
The force acting on the particle varies with time and is defined for the interval \( 0 \leq t \leq 2T \). The force is zero at \( t = T \) and reaches its maximum at \( t = 0 \) and \( t = 2T \).

2. **Finding Acceleration**:
From Newton's second law, we know that:
\[ F = M \cdot a \]
Therefore, the acceleration \( a \) can be expressed as:
\[ a = \frac{F}{M} = \frac{F_0}{M} \left( 1 - \left( \frac{t - T}{T} \right)^2 \right) \]

3. **Setting up the Integral for Velocity**:
The acceleration is the rate of change of velocity, so we can write:
\[ a = \frac{dv}{dt} \]
Thus, we have:
\[ \frac{dv}{dt} = \frac{F_0}{M} \left( 1 - \left( \frac{t - T}{T} \right)^2 \right) \]

4. **Integrating to Find Velocity**:
We need to integrate the acceleration to find the velocity:
\[ v = \int_0^{2T} \frac{F_0}{M} \left( 1 - \left( \frac{t - T}{T} \right)^2 \right) dt \]

5. **Breaking Down the Integral**:
The integral can be separated into two parts:
\[ v = \frac{F_0}{M} \left( \int_0^{2T} dt - \int_0^{2T} \left( \frac{t - T}{T} \right)^2 dt \right) \]

6. **Calculating the First Integral**:
The first integral is straightforward:
\[ \int_0^{2T} dt = 2T \]

7. **Calculating the Second Integral**:
For the second integral, we substitute \( u = \frac{t - T}{T} \), which gives us:
\[ dt = T \, du \quad \text{and the limits change from } 0 \text{ to } 2T \text{ to } -1 \text{ to } 1. \]
Thus,
\[ \int_0^{2T} \left( \frac{t - T}{T} \right)^2 dt = T^3 \int_{-1}^{1} u^2 du = T^3 \left[ \frac{u^3}{3} \right]_{-1}^{1} = T^3 \left( \frac{1}{3} - \left(-\frac{1}{3}\right) \right) = \frac{2T^3}{3} \]

8. **Combining Results**:
Now substituting back into the expression for \( v \):
\[ v = \frac{F_0}{M} \left( 2T - \frac{2T^3}{3} \right) = \frac{F_0}{M} \left( \frac{6T}{3} - \frac{2T^3}{3} \right) = \frac{F_0}{M} \cdot \frac{2T(3 - T^2)}{3} \]

9. **Final Expression for Velocity**:
Thus, the final expression for the velocity \( v \) after time \( 2T \) is:
\[ v = \frac{2F_0 T}{3M} (3 - T^2) \]