Question

A car, starting from rest, accelerates at the rate \( f \) through a distance \( S \), then continues at constant speed for some time \( t \) and then decelerates at the rate \( \frac{f}{2} \) to come to rest. If the total distance is \( 5S \), then

• A. \( S = \frac{1}{4} ft^2 \)
• B. \( S = \frac{1}{2} ft^2 \)
• C. \( S = 4ft^2 \)
• D. \( S = 2ft^2 \)

Hint:

Break motion into parts: acceleration, uniform motion, and deceleration. Then apply kinematic equations separately.

Answer 1

The Correct Option is B

Step 1: Motion during acceleration.
Initial velocity \( u = 0 \), acceleration \( f \), distance \( S \):
\[ v^2 = 2fS \Rightarrow v = \sqrt{2fS} \]

Step 2: Motion during constant velocity.
Distance covered:
\[ S_2 = v \cdot t = \sqrt{2fS} \cdot t \]

Step 3: Motion during deceleration.
Deceleration = \( \frac{f}{2} \), final velocity = 0: \[ v^2 = 2 \times \frac{f}{2} \times S_3 \Rightarrow S_3 = \frac{v^2}{f} = \frac{2fS}{f} = 2S \]

Step 4: Total distance.
\[ S_{total} = S + S_2 + S_3 = 5S \]

Step 5: Substitute values.
\[ S + \sqrt{2fS}t + 2S = 5S \]
\[ 3S + \sqrt{2fS}t = 5S \]
\[ \sqrt{2fS}t = 2S \]

Step 6: Solve equation.
Squaring both sides: \[ 2fS t^2 = 4S^2 \]
\[ 2ft^2 = 4S \Rightarrow S = \frac{1}{2} ft^2 \]

Step 7: Final Answer.
\[ \boxed{S = \frac{1}{2} ft^2} \]