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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

A car moving with a velocity of 10 m/s10 \text{ m/s} can be stopped by the application of a constant force FF in a distance of 20 m20 \text{ m}. If the velocity of the car is 30 m/s30 \text{ m/s}, it can be stopped by this force in:

A

20/3 m

B

20 m

C

60 m

D

180 m

Step-by-Step Solution

  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=ΔKW = \Delta K) or kinematic equations.
  2. Formula: Using the kinematic equation v2=u2+2asv^2 = u^2 + 2as with final velocity v=0v=0, we get the stopping distance ds=u22ad_s = \frac{-u^2}{2a} . Since the force FF is constant, the acceleration a=F/ma = F/m is constant. Therefore, dsu2d_s \propto u^2.
  3. Calculation: Case 1: u1=10 m/su_1 = 10 \text{ m/s}, d1=20 md_1 = 20 \text{ m}. Case 2: u2=30 m/su_2 = 30 \text{ m/s}. Ratio: d2d1=(u2u1)2=(3010)2=32=9\frac{d_2}{d_1} = \left(\frac{u_2}{u_1}\right)^2 = \left(\frac{30}{10}\right)^2 = 3^2 = 9. d2=9×d1=9×20 m=180 md_2 = 9 \times d_1 = 9 \times 20 \text{ m} = 180 \text{ m}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEmovingvelocitystoppedapplicationconstant

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