A body travels a distance s in t seconds. It starts from rest and ends at rest . In the first part of the journey , it moves with constant acceleration f and in the second part with constant retardation r. the value of t is given by

Options

A. `sqrt(8s ( 1/f + 1/r) )`
B. `2s (1/f + 1/r)`
C. `(2s)/(1/f + 1/r)`
D. `sqrt(2s(f + r))`

Answer

To solve the problem step by step, we will analyze the motion of the body in two parts: the first part with constant acceleration \( f \) and the second part with constant retardation \( r \).

### Step 1: Understand the motion
The body starts from rest and ends at rest. This means:
- Initial velocity \( u = 0 \)
- Final velocity \( v = 0 \) at the end of the motion.

### Step 2: Define the time intervals
Let:
- \( t_1 \) be the time taken during the acceleration phase.
- \( t_2 \) be the time taken during the retardation phase.

Given that the total time \( t = t_1 + t_2 \).

### Step 3: Use the equations of motion for the first part
For the first part of the journey (acceleration phase):
- Initial velocity \( u = 0 \)
- Acceleration \( f \)
- Time \( t_1 \)

Using the equation of motion:
\[ d_1 = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ d_1 = 0 + \frac{1}{2} f t_1^2 = \frac{1}{2} f t_1^2 \]

### Step 4: Use the equations of motion for the second part
For the second part of the journey (retardation phase):
- Initial velocity \( v = f t_1 \) (the final velocity of the first part)
- Final velocity \( v = 0 \)
- Retardation \( r \)
- Time \( t_2 \)

Using the equation of motion:
\[ d_2 = vt - \frac{1}{2} r t^2 \]
Substituting the values:
\[ d_2 = f t_1 t_2 - \frac{1}{2} r t_2^2 \]

### Step 5: Relate the distances
The total distance traveled is:
\[ s = d_1 + d_2 \]
Substituting the expressions for \( d_1 \) and \( d_2 \):
\[ s = \frac{1}{2} f t_1^2 + \left( f t_1 t_2 - \frac{1}{2} r t_2^2 \right) \]

### Step 6: Substitute \( t_2 \) in terms of \( t \)
From the total time:
\[ t_2 = t - t_1 \]

### Step 7: Substitute \( t_2 \) into the distance equation
Substituting \( t_2 \) into the distance equation:
\[ s = \frac{1}{2} f t_1^2 + f t_1 (t - t_1) - \frac{1}{2} r (t - t_1)^2 \]

### Step 8: Expand and simplify
Expanding the equation:
\[ s = \frac{1}{2} f t_1^2 + f t_1 t - f t_1^2 - \frac{1}{2} r (t^2 - 2tt_1 + t_1^2) \]
Combine like terms and simplify.

### Step 9: Rearranging the equation
Rearranging gives:
\[ s = \left( \frac{1}{2} f - f \right) t_1^2 + f t_1 t - \frac{1}{2} r t^2 + r t t_1 \]

### Step 10: Solve for \( t^2 \)
Rearranging the equation to isolate \( t^2 \):
\[ t^2 = \frac{8s}{f + r} \]
Taking the square root gives:
\[ t = \sqrt{\frac{8s}{f + r}} \]

### Final Answer
Thus, the value of \( t \) is:
\[ t = \sqrt{\frac{8s}{f + r}} \]