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 $$15S$$, then

Solution & Explanation

1. Analyze Phase 1: Acceleration

The car starts from rest, meaning its initial velocity ($$u_1$$) is zero:

$$u_1 = 0$$

It accelerates at a uniform rate $$f$$ over a distance $$S$$. Let its final velocity at the end of this phase be $$v$$. Using the third equation of motion ($$v^2 = u^2 + 2as$$):

$$v^2 = 0^2 + 2 \cdot f \cdot S$$

$$v^2 = 2 \cdot f \cdot S \quad \text{--- (Eq. 1)}$$

2. Analyze Phase 2: Constant Speed

The car continues to travel at this constant maximum speed $$v$$ for a time duration $$t$$. Let the distance covered during this phase be $$S_2$$:

$$S_2 = v \cdot t \quad \text{--- (Eq. 2)}$$

3. Analyze Phase 3: Deceleration

The car then decelerates at a uniform rate of $$\frac{f}{2}$$ to come completely to rest. Let us define the parameters for this final phase:

• Initial velocity: $$v$$
• Final velocity: $$0$$
• Deceleration: $$a_3 = -\frac{f}{2}$$
• Distance traveled: Let this be $$S_3$$

Applying the third equation of motion again:

$$0^2 = v^2 + 2 \cdot \left(-\frac{f}{2}\right) \cdot S_3$$

$$0 = v^2 - f \cdot S_3$$

$$S_3 = \frac{v^2}{f}$$

Substituting the value of $$v^2 = 2 \cdot f \cdot S$$ from Equation 1 into this expression gives:

$$S_3 = \frac{2 \cdot f \cdot S}{f} = 2S$$

4. Set Up the Total Distance Equation

The total distance traversed by the car during all three phases of its journey combined is given as $$15S$$:

$$S_{\text{total}} = S + S_2 + S_3 = 15S$$

Substitute our known values for $$S_3 = 2S$$ into this relation:

$$S + S_2 + 2S = 15S$$

$$3S + S_2 = 15S$$

$$S_2 = 12S$$

5. Relate $$S$$, $$f$$, and $$t$$

Substitute the expression for $$S_2$$ from Equation 2 ($$S_2 = v \cdot t$$) into our new relation:

$$v \cdot t = 12S$$

Square both sides of the equation to easily substitute our initial $$v^2$$ term:

$$v^2 \cdot t^2 = 144 \cdot S^2$$

Substitute $$v^2 = 2 \cdot f \cdot S$$ from Equation 1:

$$(2 \cdot f \cdot S) \cdot t^2 = 144 \cdot S^2$$

Divide both sides by $$2S$$ (since distance $$S \neq 0$$):

$$f \cdot t^2 = 72 \cdot S$$

Isolating the value of distance $$S$$ yields:

$$S = \frac{1}{72} \cdot f \cdot t^2$$

Concept Check: Because this specific mathematical relation is missing from options A, B, and C, the correct choice is "None of these".

Correct Option Key: None of these