A car moving with a speed of $40 ext{ km/h}$ can be stopped by applying brakes for atleast $2 ext{ m}$. If

Question

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

Accepted Answer

1.  **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 ($F \cdot d \propto v^2$) .
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)" .
3.  **Calculation:**
    *   Initial speed $v_1 = 40 	ext{ km/h}$, stopping distance $d_1 = 2 	ext{ m}$.
    *   New speed $v_2 = 80 	ext{ km/h}$ (which is $2 	imes v_1$).
    *   New stopping distance $d_2 = 4 	imes d_1 = 4 	imes 2 	ext{ m} = 8 	ext{ m}$.

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