A car moving with a velocity of $10 ext{ m/s}$ can be stopped by the application of a constant force $F$ in

Question

A car moving with a velocity of $10 	ext{ m/s}$ can be stopped by the application of a constant force $F$ in a distance of $20 	ext{ m}$. If the velocity of the car is $30 	ext{ m/s}$, it can be stopped by this force in:

Accepted Answer

1.  **Concept:** The stopping distance of a vehicle is directly proportional to the square of its initial velocity, provided the retarding force (and thus deceleration) remains constant. This can be derived from the work-energy theorem ($W = \Delta K$) or kinematic equations.
2.  **Formula:** Using the kinematic equation $v^2 = u^2 + 2as$ with final velocity $v=0$, we get the stopping distance $d_s = \frac{-u^2}{2a}$ . Since the force $F$ is constant, the acceleration $a = F/m$ is constant. Therefore, $d_s \propto u^2$.
3.  **Calculation:**
    *   Case 1: $u_1 = 10 	ext{ m/s}$, $d_1 = 20 	ext{ m}$.
    *   Case 2: $u_2 = 30 	ext{ m/s}$.
    *   Ratio: $\frac{d_2}{d_1} = \left(\frac{u_2}{u_1}ight)^2 = \left(\frac{30}{10}ight)^2 = 3^2 = 9$.
    *   $d_2 = 9 	imes d_1 = 9 	imes 20 	ext{ m} = 180 	ext{ m}$.

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