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Step-by-Step Solution
Step 1: Identify the three phases of motion
1. The car starts from rest and accelerates with acceleration $f$ over a distance $S$.
2. It then continues to move at the constant speed (just reached) for a time $t$.
3. Finally, it decelerates at the rate $f/2$ until it comes to rest.
Step 2: Find the speed after the first phase
Since the car starts from rest and accelerates with acceleration $f$ through distance $S$, we use the equation of motion:
$v^2 = u^2 + 2 a s.$
Here, $u = 0$, $a = f$, and $s = S$. Thus,
$v^2 = 2 f S \quad \Rightarrow \quad v = \sqrt{2 f S}.$
Step 3: Distance traveled during the constant speed phase
The car moves at speed $v = \sqrt{2 f S}$ for a time $t$. The distance covered in this phase is:
Distance $= v \times t = \sqrt{2 f S} \times t.$
Step 4: Calculate the distance covered during deceleration
The car decelerates from speed $v = \sqrt{2 f S}$ to rest with deceleration $f/2$. Using
$v^2 = u^2 - 2 a s,$
where final velocity $= 0$, initial velocity $= \sqrt{2 f S}$, and $a = f/2$ (negative sign taken care of in the distance calculation), we get:
$0 = ( \sqrt{2 f S} )^2 - 2 \left(\frac{f}{2}\right) S_2.$
Simplifying:
$0 = 2 f S - f S_2 \quad \Rightarrow \quad S_2 = 2 S.$
Hence, the distance traveled while decelerating is $2 S$.
Step 5: Express the total distance covered
According to the problem statement, the total distance traveled is $15 S$.
So, summing up all three parts:
$S\ (\text{acceleration}) \;+\; \sqrt{2 f S}\,t\ (\text{constant speed}) \;+\; 2S\ (\text{deceleration}) \;=\; 15 S.$
Hence,
$S + \sqrt{2 f S}\,t + 2S = 15S.$
This simplifies to:
$\sqrt{2 f S}\,t = 12 S.$
Step 6: Solve for $S$ in terms of $f$ and $t$
From $\sqrt{2 f S}\,t = 12 S,$ we get
$\sqrt{2 f S} = \frac{12 S}{t}.$
Squaring both sides,
$2 f S = \frac{144 S^2}{t^2} \quad \Rightarrow \quad f = \frac{144 S^2}{2 S t^2} \quad \Rightarrow \quad f = \frac{72 S}{t^2}.$
Therefore,
$S = \frac{f t^2}{72}.$
In simpler form:
$S = \frac{1}{72} f t^2.$
Final Answer
$S = \frac{1}{72} f t^2.$
