A car, starting from rest, accelerates at the rate (f) through a distance (S), then continues at constant speed for some time (t) and then decelerates at the rate ` f//2` to come to rest. If the total distance is ` 5 S`, then prove that
` S=1/2/ ft^(2)`.

Explanation

To solve the problem, we will break it down into three parts based on the motion of the car: acceleration, constant speed, and deceleration.

### Step-by-Step Solution:

1. **Acceleration Phase:**
- The car starts from rest and accelerates at a rate \( f \) through a distance \( S \).
- Using the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \) (initial velocity), \( a = f \) (acceleration), and \( S = s \) (distance).
- Thus, we have:
\[ s = 0 + \frac{1}{2} f t_1^2 \quad \text{(where \( t_1 \) is the time of acceleration)} \]
- This simplifies to:
\[ s = \frac{1}{2} f t_1^2 \quad \text{(Equation 1)} \]

2. **Constant Speed Phase:**
- After accelerating, the car continues at a constant speed for a time \( t \).
- The speed at the end of the acceleration phase is:
\[ v = f t_1 \quad \text{(Equation 2)} \]
- The distance covered during the constant speed phase \( S_2 \) is:
\[ S_2 = v \cdot t = (f t_1) \cdot t = f t_1 t \quad \text{(Equation 3)} \]

3. **Deceleration Phase:**
- The car then decelerates at a rate of \( \frac{f}{2} \) until it comes to rest.
- Using the equation of motion for this phase:
\[ 0 = v^2 - 2 a s_3 \]
where \( v = f t_1 \) (initial speed), \( a = \frac{f}{2} \) (deceleration), and \( s_3 \) is the distance during deceleration.
- Rearranging gives:
\[ (f t_1)^2 = 2 \cdot \frac{f}{2} \cdot s_3 \]
Simplifying:
\[ f^2 t_1^2 = f s_3 \quad \Rightarrow \quad s_3 = f t_1^2 \quad \text{(Equation 4)} \]

4. **Total Distance:**
- The total distance covered by the car is given as \( 5S \):
\[ S + S_2 + s_3 = 5S \]
- Substituting the equations we derived:
\[ s + f t_1 t + f t_1^2 = 5s \]
- Replacing \( s \) from Equation 1:
\[ \frac{1}{2} f t_1^2 + f t_1 t + f t_1^2 = 5 \left(\frac{1}{2} f t_1^2\right) \]
- Simplifying the left side:
\[ \frac{1}{2} f t_1^2 + f t_1 t + f t_1^2 = \frac{1}{2} f t_1^2 + \frac{2}{2} f t_1^2 + f t_1 t = \frac{5}{2} f t_1^2 + f t_1 t \]
- Setting the equation:
\[ \frac{5}{2} f t_1^2 + f t_1 t = \frac{5}{2} f t_1^2 \]
- This implies:
\[ f t_1 t = 0 \]
- Since \( f \) and \( t_1 \) cannot be zero, we can conclude that \( t_1 = t \).

5. **Final Result:**
- Substituting \( t_1 = t \) back into Equation 1:
\[ s = \frac{1}{2} f t^2 \]
- Thus, we have proved that:
\[ S = \frac{1}{2} f t^2 \]