A car , starting from rest, accelerates at the rate `f` through a distance `S` then continues at constant speed for time `t` and then decelerates at the rate `(f)/(2)` to come to rest . If the total distance traversed is `15 S` , then

Options

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

Explanation

To solve the problem step by step, let's break down the motion of the car into three distinct parts: acceleration, constant speed, and deceleration.

### Step 1: Analyze the first part of the motion (Acceleration)
The car starts from rest and accelerates at a rate of \( f \) through a distance \( S \).

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 0 \) (initial velocity),
- \( a = f \) (acceleration),
- \( s = S \) (distance).

Substituting the values:
\[ v^2 = 0 + 2fS \]
Thus, we find:
\[ v = \sqrt{2fS} \]

### Step 2: Analyze the second part of the motion (Constant Speed)
In this part, the car travels at a constant speed \( v \) for a time \( t \). The distance covered during this time is:
\[ x = vt \]
Substituting the value of \( v \):
\[ x = \sqrt{2fS} \cdot t \]

### Step 3: Analyze the third part of the motion (Deceleration)
The car decelerates at a rate of \( \frac{f}{2} \) until it comes to rest. The initial velocity for this part is \( v \) and the final velocity is \( 0 \).

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( u = v = \sqrt{2fS} \),
- \( a = -\frac{f}{2} \) (deceleration),
- \( s = y \) (distance during deceleration).

Substituting the values:
\[ 0 = (2fS) + 2\left(-\frac{f}{2}\right)y \]
This simplifies to:
\[ 0 = 2fS - fy \]
Rearranging gives:
\[ fy = 2fS \implies y = 2S \]

### Step 4: Total Distance
The total distance traveled by the car is given as \( 15S \). Therefore, we can write:
\[ S + x + y = 15S \]
Substituting \( x \) and \( y \):
\[ S + \sqrt{2fS} \cdot t + 2S = 15S \]
This simplifies to:
\[ 3S + \sqrt{2fS} \cdot t = 15S \]
Rearranging gives:
\[ \sqrt{2fS} \cdot t = 12S \]

### Step 5: Solve for \( S \)
Now, squaring both sides:
\[ 2fS \cdot t^2 = 144S^2 \]
Dividing both sides by \( S \) (assuming \( S \neq 0 \)):
\[ 2ft^2 = 144S \]
Thus, we can express \( S \) as:
\[ S = \frac{2ft^2}{144} = \frac{1}{72} ft^2 \]

### Final Answer
The value of \( S \) is:
\[ S = \frac{1}{72} ft^2 \]