A particle of mass m, initially at rest, is acted on by a force `F=F_0{1-((2t-T)/T)^2}` during the interval `0lt tlt T`. The velocity of the particle at the end of the interval is :

Options

A. `(5F_0T)/(6m)`
B. `(4F_0T)/(3m)`
C. `(2F_0T)/(3m)`
D. `(3F_0T)/(2m)`

Answer

To find the velocity of a particle acted upon by a time-dependent force, we can follow these steps:

### Step 1: Understand the Given Information
We have a particle of mass \( m \) that is initially at rest. The force acting on the particle is given by:
\[ F = F_0 \left( 1 - \left( \frac{2t - T}{T} \right)^2 \right) \]
This force is valid for the time interval \( 0 < t < T \).

### Step 2: Apply Newton's Second Law
According to Newton's second law, the force acting on an object is equal to the mass of the object multiplied by its acceleration:
\[ F = m \cdot a \]
Since acceleration \( a \) is the rate of change of velocity, we can express it as:
\[ a = \frac{dv}{dt} \]
Thus, we can rewrite the equation as:
\[ F = m \frac{dv}{dt} \]

### Step 3: Rearrange the Equation
We can rearrange the equation to isolate \( dv \):
\[ m \, dv = F \, dt \]

### Step 4: Integrate Both Sides
Now, we integrate both sides. The left side will be integrated with respect to \( v \) from \( 0 \) to \( v \) (the final velocity), and the right side will be integrated with respect to \( t \) from \( 0 \) to \( T \):
\[ \int_0^v m \, dv = \int_0^T F \, dt \]
This simplifies to:
\[ mv = \int_0^T F \, dt \]

### Step 5: Substitute the Expression for Force
Now we substitute the expression for the force \( F \):
\[ mv = \int_0^T F_0 \left( 1 - \left( \frac{2t - T}{T} \right)^2 \right) dt \]

### Step 6: Simplify the Integral
We can break this integral into two parts:
\[ mv = \int_0^T F_0 \, dt - \int_0^T F_0 \left( \frac{2t - T}{T} \right)^2 dt \]
Calculating the first integral:
\[ \int_0^T F_0 \, dt = F_0 T \]

For the second integral, we simplify:
\[ \int_0^T \left( \frac{2t - T}{T} \right)^2 dt = \frac{4}{T^2} \int_0^T (t^2 - 2tT + \frac{T^2}{4}) dt \]
Calculating this integral gives:
\[ \int_0^T t^2 dt = \frac{T^3}{3}, \quad \int_0^T t dt = \frac{T^2}{2} \]
Thus, the second integral evaluates to:
\[ \frac{4}{T^2} \left( \frac{T^3}{3} - 2T \cdot \frac{T^2}{2} + \frac{T^3}{4} \right) = \frac{4}{T^2} \left( \frac{T^3}{3} - T^3 + \frac{T^3}{4} \right) \]
Combining these terms gives:
\[ \frac{4}{T^2} \left( \frac{4T^3}{12} - \frac{12T^3}{12} + \frac{3T^3}{12} \right) = \frac{4}{T^2} \left( -\frac{5T^3}{12} \right) = -\frac{20T}{12} = -\frac{5T}{3} \]

### Step 7: Combine Results
Putting it all together:
\[ mv = F_0 T - \left( -\frac{5F_0 T}{3} \right) = F_0 T + \frac{5F_0 T}{3} = \frac{8F_0 T}{3} \]

### Step 8: Solve for Velocity
Finally, we solve for \( v \):
\[ v = \frac{8F_0 T}{3m} \]

### Final Answer
The velocity of the particle at the end of the interval is:
\[ v = \frac{8F_0 T}{3m} \]