A particle moving along x-axis has acceleration `f`, at time `t`, given by `f = f_0 (1 - (t)/(T))`, where `f_0` and `T` are constant.
The particle at `t = 0` has zero velocity. In the time interval between `t = 0` and the instant when `f = 0`, the particle's velocity `(v_x)` is :

Options

A. `f_0 T`
B. `(1)/(2) f_0 T^2`
C. `f_0 T^2`
D. `(1)/(2) f_0 T`

Explanation

To solve the problem, we need to find the velocity of a particle moving along the x-axis with a given acceleration function. Here are the steps to derive the solution:

### Step 1: Understand the given acceleration function
The acceleration \( f \) is given by:
\[ f = f_0 \left(1 - \frac{t}{T}\right) \]
where \( f_0 \) and \( T \) are constants.

### Step 2: Set up the relationship between acceleration and velocity
Acceleration is the derivative of velocity with respect to time:
\[ f = \frac{dv}{dt} \]
Substituting the expression for \( f \):
\[ \frac{dv}{dt} = f_0 \left(1 - \frac{t}{T}\right) \]

### Step 3: Integrate the acceleration to find velocity
To find the velocity \( v \), we integrate the acceleration with respect to time:
\[ dv = f_0 \left(1 - \frac{t}{T}\right) dt \]
Integrating both sides from \( t = 0 \) to \( t \) and \( v = 0 \) to \( v \):
\[ \int_0^v dv = f_0 \int_0^t \left(1 - \frac{t}{T}\right) dt \]
This gives:
\[ v = f_0 \left( t - \frac{t^2}{2T} \right) \]

### Step 4: Find the time when acceleration becomes zero
Set the acceleration \( f \) to zero to find the time \( t \) when this occurs:
\[ 0 = f_0 \left(1 - \frac{t}{T}\right) \]
Solving for \( t \):
\[ 1 - \frac{t}{T} = 0 \implies t = T \]

### Step 5: Substitute \( t = T \) into the velocity equation
Now substitute \( t = T \) back into the velocity equation:
\[ v = f_0 \left( T - \frac{T^2}{2T} \right) \]
This simplifies to:
\[ v = f_0 \left( T - \frac{T}{2} \right) = f_0 \left( \frac{T}{2} \right) \]

### Final Result
Thus, the velocity of the particle at the instant when \( f = 0 \) is:
\[ v = \frac{f_0 T}{2} \]