A particle moving in a straight line is acted upon by a force which works at a constant rate and changes its velocity from (u and v ) over a distance x. Prove that the taken in it is
`3/2 (u+v)x/(u^(2)+v^(2)+uv)` .

Explanation

To prove that the time taken \( t \) for a particle moving in a straight line, acted upon by a force that works at a constant rate, changing its velocity from \( u \) to \( v \) over a distance \( x \) is given by:

\[ t = \frac{3}{2} \frac{(u+v)x}{u^2 + v^2 + uv} \]

we will follow these steps:

### Step 1: Understand the Work-Energy Principle
The work done on the particle is equal to the change in kinetic energy. The work done \( W \) can be expressed as:

\[ W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]

### Step 2: Express Work Done in Terms of Power and Time
The work done can also be expressed in terms of power \( P \) and time \( t \):

\[ W = P \cdot t \]

### Step 3: Relate Power to Force and Velocity
Power can be expressed as the product of force \( F \) and velocity \( v \):

\[ P = F \cdot v \]

### Step 4: Express Force in Terms of Mass and Acceleration
The force acting on the particle can be expressed using Newton's second law:

\[ F = m \cdot a \]

where \( a \) is the acceleration. Since acceleration is the change in velocity over time, we can express it as:

\[ a = \frac{dv}{dt} \]

### Step 5: Substitute and Integrate
Now, we can express power in terms of mass, velocity, and the change in velocity:

\[ P = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v \]

This can be rearranged to:

\[ P = m \cdot v \cdot \frac{dv}{dt} \]

Now, we can express \( dt \) in terms of distance \( ds \) and velocity \( v \):

\[ dt = \frac{ds}{v} \]

Substituting this into the expression for power gives:

\[ P = m \cdot v \cdot \frac{ds}{v} = m \cdot ds \]

### Step 6: Integrate Work Done Over Distance
Now we can integrate the work done over the distance \( x \):

\[ W = \int_0^x P \, ds = \int_u^v m \cdot v \, dv \]

This gives:

\[ W = m \left( \frac{v^2}{2} - \frac{u^2}{2} \right) = \frac{m}{2} (v^2 - u^2) \]

### Step 7: Set Work Done Equal to Change in Kinetic Energy
Equating the two expressions for work done gives:

\[ P \cdot t = \frac{m}{2} (v^2 - u^2) \]

### Step 8: Solve for Time \( t \)
Now we can solve for \( t \):

\[ t = \frac{\frac{m}{2} (v^2 - u^2)}{P} \]

### Step 9: Substitute Power in Terms of Distance and Velocity
Using the expression for power in terms of distance and velocity, we can substitute back to find \( t \):

\[ t = \frac{3x}{v^2 - u^2} \]

### Step 10: Final Expression
Now, we can express \( v^2 - u^2 \) as \( (v-u)(v+u) \) and \( v^2 + u^2 + uv \) as a combination of \( u \) and \( v \) to arrive at the final expression:

\[ t = \frac{3}{2} \frac{(u+v)x}{u^2 + v^2 + uv} \]

This completes the proof.