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)`.

Options

A. ` S = (1)/(6) ft^(2)`
B. ` S = ft`
C. ` s = (1)/(4) ft^(2)`
D. ` s = (1)/(72) ft^(2)`

Explanation

To solve this problem, we need to analyze the motion of the car in three phases: acceleration, constant speed, and deceleration. Let's break down each phase step by step.

### Step 1: Acceleration Phase
The car starts from rest and accelerates at a rate \( f \) through a distance \( S \).

- Initial velocity (\( u \)) = 0
- Acceleration (\( a \)) = \( f \)
- Distance (\( S \))

Using the equation of motion:
\[ v^2 = u^2 + 2aS \]
\[ v_1^2 = 0 + 2fS \]
\[ v_1 = \sqrt{2fS} \]

### Step 2: Constant Speed Phase
The car continues at a constant speed \( v_1 \) for a time \( t \).

- Speed (\( v_1 \)) = \( \sqrt{2fS} \)
- Time (\( t \))

Distance covered during this phase (\( S_2 \)):
\[ S_2 = v_1 \cdot t \]
\[ S_2 = \sqrt{2fS} \cdot t \]

### Step 3: Deceleration Phase
The car decelerates at a rate \( \frac{f}{2} \) to come to rest.

- Initial velocity (\( u \)) = \( v_1 \)
- Final velocity (\( v_f \)) = 0
- Deceleration (\( a \)) = \( -\frac{f}{2} \)

Using the equation of motion:
\[ v_f^2 = u^2 + 2aS \]
\[ 0 = v_1^2 - 2 \left( \frac{f}{2} \right) S_3 \]
\[ 0 = 2fS - fS_3 \]
\[ S_3 = 2S \]

### Total Distance
The total distance covered by the car is given as \( 5S \):
\[ S + S_2 + S_3 = 5S \]
\[ S + \sqrt{2fS} \cdot t + 2S = 5S \]
\[ 3S + \sqrt{2fS} \cdot t = 5S \]
\[ \sqrt{2fS} \cdot t = 2S \]
\[ t = \frac{2S}{\sqrt{2fS}} \]
\[ t = \frac{2\sqrt{S}}{\sqrt{2f}} \]
\[ t = \sqrt{\frac{2S}{f}} \]

### Proving \( S = \frac{1}{2} ft^2 \)
Now, we need to express \( S \) in terms of \( f \) and \( t \):
\[ t^2 = \frac{2S}{f} \]
\[ S = \frac{1}{2} ft^2 \]

Thus, we have proven that:
\[ S = \frac{1}{2} ft^2 \]