Question

If the initial speed of the car moving at constant acceleration is halved, then the stopping distance \( S \) becomes:

• A. \( 2S \)
• B. \( \frac{S}{2} \)
• C. \( 4S \)
• D. \( \frac{S}{4} \)
• E. \( \frac{S}{8} \)

Hint:

In problems involving stopping distance with constant acceleration, the stopping distance is proportional to the square of the initial speed. Halving the initial speed results in a reduction of the stopping distance by a factor of 4.

Answer 1

The Correct Option is D

The stopping distance \( S \) of an object moving with initial speed \( u \) and constant acceleration \( a \) can be given by the kinematic equation: \[ v^2 = u^2 + 2aS \] 
where: 
- \( v \) is the final speed (which is 0 when the car stops), 
- \( u \) is the initial speed, 
- \( a \) is the acceleration (negative because the car is decelerating), 
- \( S \) is the stopping distance. 
Since \( v = 0 \) (the car stops), we can rewrite the equation as: \[ 0 = u^2 + 2aS \quad \Rightarrow \quad S = \frac{u^2}{-2a} \] 
Now, if the initial speed \( u \) is halved, the new initial speed becomes \( \frac{u}{2} \). Substituting \( \frac{u}{2} \) into the equation for the new stopping distance \( S' \), we get: \[ S' = \frac{\left(\frac{u}{2}\right)^2}{-2a} = \frac{u^2}{4(-2a)} = \frac{S}{4} \] Thus, when the initial speed is halved, the stopping distance becomes \( \frac{S}{4} \).
Thus, the correct answer is option (D), \( \frac{S}{4} \).