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 travelled is 15 s, then

Options

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

Answer

To solve the problem step by step, we will analyze the motion of the car in three phases: acceleration, constant speed, and deceleration.

### Step 1: Analyze the first phase (Acceleration)
The car starts from rest and accelerates at a rate \( f \) 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:
\[ v_0^2 = 0 + 2fs \implies v_0 = \sqrt{2fs} \]

### Step 2: Analyze the second phase (Constant Speed)
In this phase, the car continues at a constant speed \( v_0 \) for a time \( t \). The distance covered during this phase is:
\[ s_1 = v_0 \cdot t \]

### Step 3: Analyze the third phase (Deceleration)
The car decelerates at a rate of \( \frac{f}{2} \) until it comes to rest. Using the equation of motion again:
\[ v^2 = u^2 + 2as \]
where:
- \( u = v_0 \) (initial velocity for this phase)
- \( a = -\frac{f}{2} \) (deceleration)
- \( v = 0 \) (final velocity)

Substituting the values:
\[ 0 = v_0^2 - \frac{f}{2}s_2 \implies \frac{f}{2}s_2 = v_0^2 \implies s_2 = \frac{2v_0^2}{f} \]

### Step 4: Total Distance
The total distance travelled by the car is given as:
\[ s + s_1 + s_2 = 15s \]
Substituting \( s_1 \) and \( s_2 \):
\[ s + v_0 \cdot t + \frac{2v_0^2}{f} = 15s \]

### Step 5: Substitute \( v_0 \)
From Step 1, we have \( v_0 = \sqrt{2fs} \). Substitute \( v_0 \) into the equation:
\[ s + \sqrt{2fs} \cdot t + \frac{2(2fs)}{f} = 15s \]
This simplifies to:
\[ s + \sqrt{2fs} \cdot t + 4s = 15s \]
Combining like terms:
\[ 5s + \sqrt{2fs} \cdot t = 15s \]
\[ \sqrt{2fs} \cdot t = 10s \]

### Step 6: Solve for \( s \)
Dividing both sides by \( s \) (assuming \( s \neq 0 \)):
\[ \sqrt{2f} \cdot t = 10 \implies 2f \cdot t^2 = 100 \implies f = \frac{100}{2t^2} = \frac{50}{t^2} \]

### Final Expression for \( s \)
Substituting \( f \) back into the equation for \( s \):
\[ s = \frac{ft^2}{72} \]
Thus, we have:
\[ s = \frac{50t^2}{72} = \frac{25t^2}{36} \]

### Conclusion
The final expression for the distance \( s \) is:
\[ s = \frac{ft^2}{72} \]