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 = (1)/(72) ft^(2)`
B. `s = (1)/(36) ft^(2)`
C. `s = (1)/(2) ft^(2)`
D. `s = (1)/(4) ft^(2)`

Answer

To solve the problem step by step, we will break down the motion of the car into three distinct phases: acceleration, constant speed, and deceleration. We will derive the relationships between the distance traveled, time, and acceleration.

### 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:
\[ v_1^2 = 0 + 2fs \implies v_1 = \sqrt{2fs} \]
This gives us the velocity \( v_1 \) at the end of the acceleration phase.

### Step 2: Analyze the second phase (Constant Speed)
The car then travels at constant speed \( v_1 \) for a time \( t \). The distance covered during this phase is:
\[ s_2 = v_1 \cdot t = \sqrt{2fs} \cdot t \]

### Step 3: Analyze the third phase (Deceleration)
The car decelerates at a rate of \( \frac{f}{2} \) until it comes to rest. The initial velocity for this phase is \( v_1 \), and the final velocity is \( 0 \).

Using the equation of motion again:
\[ v^2 = u^2 + 2as \]
where:
- \( u = v_1 \),
- \( v = 0 \),
- \( a = -\frac{f}{2} \).

Substituting the values:
\[ 0 = v_1^2 - 2 \left(-\frac{f}{2}\right) s_3 \]
This simplifies to:
\[ v_1^2 = fs_3 \implies s_3 = \frac{v_1^2}{f} \]

### Step 4: Substitute \( v_1 \)
From Step 1, we found \( v_1 = \sqrt{2fs} \). Substituting this into the equation for \( s_3 \):
\[ s_3 = \frac{(\sqrt{2fs})^2}{f} = \frac{2fs}{f} = 2s \]

### Step 5: Total distance traveled
The total distance traveled by the car is given as:
\[ s + s_2 + s_3 = 15s \]
Substituting \( s_2 \) and \( s_3 \):
\[ s + \sqrt{2fs} \cdot t + 2s = 15s \]
This simplifies to:
\[ 3s + \sqrt{2fs} \cdot t = 15s \]
Rearranging gives:
\[ \sqrt{2fs} \cdot t = 12s \]

### Step 6: Solve for \( t \)
Dividing both sides by \( s \) (assuming \( s \neq 0 \)):
\[ \sqrt{2f} \cdot t = 12 \implies t = \frac{12}{\sqrt{2f}} = \frac{12\sqrt{2}}{2\sqrt{f}} = \frac{6\sqrt{2}}{\sqrt{f}} \]

### Step 7: Relate \( s \) and \( t \)
Now, squaring both sides to eliminate the square root:
\[ t^2 = \frac{72}{f} \cdot s \]
Thus, we can express \( s \) in terms of \( t^2 \):
\[ s = \frac{f}{72} t^2 \]

### Final Answer
The relationship between \( s \) and \( t \) is:
\[ s = \frac{f}{72} t^2 \]