Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A car moving with a speed of $40 \text{ km/h}$ can be stopped by applying brakes for atleast $2 \text{ m}$ . If the same car is moving with a speed of $80 \text{ km/h}$ , what is the minimum stopping distance?

8 m

2 m

4 m

6 m

Step-by-Step Solution

Concept: The stopping distance ( $d_s$ ) of a vehicle is proportional to the square of the initial velocity ( $v_0$ ) provided the deceleration ( $a$ ) remains constant. This is derived from the kinematic equation $v^2 = v_0^2 + 2ax$ (where final velocity $v=0$ ) giving $d_s = \frac{-v_0^2}{2a}$ . Alternatively, using the Work-Energy Theorem, the work done by the braking force equals the change in kinetic energy ( $F \cdot d \propto v^2$ ) .
Direct Application: The NCERT text explicitly states: "Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration)" .
Calculation: Initial speed $v_1 = 40 \text{ km/h}$ , stopping distance $d_1 = 2 \text{ m}$ . New speed $v_2 = 80 \text{ km/h}$ (which is $2 \times v_1$ ).

New stopping distance $d_2 = 4 \times d_1 = 4 \times 2 \text{ m} = 8 \text{ m}$ .

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