Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

20/3 m

20 m

60 m

180 m

Step-by-Step Solution

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.
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, $d_s \propto u^2$ .
Calculation: Case 1: $u_1 = 10 \text{ m/s}$ , $d_1 = 20 \text{ m}$ . Case 2: $u_2 = 30 \text{ m/s}$ . Ratio: $\frac{d_2}{d_1} = \left(\frac{u_2}{u_1}\right)^2 = \left(\frac{30}{10}\right)^2 = 3^2 = 9$ . $d_2 = 9 \times d_1 = 9 \times 20 \text{ m} = 180 \text{ m}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started