Two particles move in the same straight line starting at the same moment from the same point in the same direction. The first moves with constant velocity u and the second starts from rest with constant acceleralion f . Then

Options

A. they will be at he greatest distance at the end of time `(u)/(2f)` from the start
B. they will be at the greatest distance at the end of time `(u)/(f)` from the start
C. their greatest distance is `(u^(2))/(2f)`
D. their greatest distance is `(u^(2))/(f)`

Answer

To solve the problem, we need to analyze the motion of two particles moving in the same straight line.

### Step-by-Step Solution:

1. **Define the Motion of the First Particle:**
The first particle moves with a constant velocity \( u \). The distance \( x_1 \) covered by the first particle after time \( t \) is given by:
\[ x_1 = ut \]

2. **Define the Motion of the Second Particle:**
The second particle starts from rest and moves with a constant acceleration \( f \). The distance \( x_2 \) covered by the second particle after time \( t \) is given by:
\[ x_2 = \frac{1}{2} f t^2 \]

3. **Determine the Distance Between the Two Particles:**
The distance \( s \) between the two particles at time \( t \) is:
\[ s = x_1 - x_2 = ut - \frac{1}{2} f t^2 \]

4. **Maximize the Distance \( s \):**
To find the greatest distance, we need to differentiate \( s \) with respect to \( t \) and set the derivative to zero:
\[ \frac{ds}{dt} = u - ft \]
Setting this equal to zero gives:
\[ u - ft = 0 \implies t = \frac{u}{f} \]

5. **Substitute \( t \) Back into the Distance Equation:**
Now, substitute \( t = \frac{u}{f} \) back into the equation for \( s \):
\[ s = u\left(\frac{u}{f}\right) - \frac{1}{2} f \left(\frac{u}{f}\right)^2 \]
Simplifying this:
\[ s = \frac{u^2}{f} - \frac{1}{2} f \cdot \frac{u^2}{f^2} = \frac{u^2}{f} - \frac{u^2}{2f} = \frac{u^2}{f} - \frac{u^2}{2f} \]
\[ = \frac{2u^2}{2f} - \frac{u^2}{2f} = \frac{u^2}{2f} \]

6. **Conclusion:**
The greatest distance between the two particles is:
\[ s = \frac{u^2}{2f} \]