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:

Use the motion equation relating speed, deceleration and stopping distance.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Stopping distance with constant deceleration: \(0 = u^2 - 2as\), so \(s = \dfrac{u^2}{2a}\).

Step 2: Equal Distance Condition
First case: \(s = \dfrac{V^2}{2a}\). Second case with speed \(nV\) and deceleration \(a'\): \(s = \dfrac{n^2V^2}{2a'}\).

Step 3: Calculation
Equating the distances, \(\dfrac{V^2}{2a} = \dfrac{n^2V^2}{2a'}\), so \(a' = n^2 a\).

Step 4: Compare the options
The deceleration needed is \(n^2 \cdot a\), the third option.

Final Answer:
The deceleration must be n squared times a. \[ \boxed{\text{(C) }n^{2}\cdot a} \]