Step 1: Understanding the Question:
The problem asks for the required deceleration to stop a car from a new speed \( nV \) within the original stopping distance \( s \). Note: The "same time" condition is mathematically inconsistent for constant deceleration if both \( V \) and \( s \) are fixed, so the primary constraint usually considered in such competitive exams is the "same distance".
Step 2: Key Formula or Approach:
From Newton's third equation of motion:
\[ v^2 = u^2 - 2as \]
To stop, final velocity \( v = 0 \):
\[ 0 = u^2 - 2as \implies a = \frac{u^2}{2s} \]
Step 3: Detailed Explanation:
Initial Case: speed = \( V \), deceleration = \( a \), distance = \( s \).
\[ a = \frac{V^2}{2s} \]
New Case: speed = \( nV \), distance = \( s \), new deceleration = \( a' \).
\[ a' = \frac{(nV)^2}{2s} \]
\[ a' = n^2 \frac{V^2}{2s} \]
Substituting the value of \( a \) from the initial case:
\[ a' = n^2 a \]
Step 4: Final Answer:
The deceleration must be \( n^2 a \).