Question

A car moving at a speed 'V' is stopped in a certain distance with deceleration 'a'. If the speed of the car is 'nV', what must be the deceleration of the car to stop it in the same distance and in the same time?

• A. $\sqrt{n} \cdot a$
• B. $n \cdot a$
• C. $n^{2} \cdot a$
• D. $n^{3} \cdot a$

Hint:

Stopping distance $s \propto u^2/a$. Stopping time $t \propto u/a$.

Answer 1

The Correct Option is B

Step 1: Formula
From kinematic equations: $v = u + at$ and $v^2 = u^2 + 2as$.

Step 2: Analysis
For stopping ($v=0$): $0 = u - at \implies a = u/t$.
Also, $0 = u^2 - 2as \implies a = u^2/2s$.

Step 3: Calculation
To stop in the same time $t$: $a' = (nu)/t = n(u/t) = na$.
To stop in the same distance $s$: $a' = (nu)^2/2s = n^2(u^2/2s) = n^2a$.
Note: The question asks for both same distance and same time, which is only possible if $n=1$. However, in standard MHT-CET variants focusing on time, $a \propto u$.

Step 4: Conclusion
Based on the available options and the "same time" constraint, $a' = n \cdot a$. Final Answer: (B)