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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

The position xx of a particle varies with time tt as x=at2bt3x = at^2 - bt^3. The acceleration of the particle will be zero at time tt equal to:

A

ab\frac{a}{b}

B

2a3b\frac{2a}{3b}

C

a3b\frac{a}{3b}

D

zero

Step-by-Step Solution

  1. Velocity (vv): Instantaneous velocity is defined as the rate of change of position with respect to time, v=dxdtv = \frac{dx}{dt} . Given x=at2bt3x = at^2 - bt^3. Differentiating with respect to tt: v=ddt(at2bt3)=2at3bt2v = \frac{d}{dt}(at^2 - bt^3) = 2at - 3bt^2.
  2. Acceleration (aacca_{acc}): Instantaneous acceleration is defined as the rate of change of velocity with respect to time, aacc=dvdta_{acc} = \frac{dv}{dt} . Differentiating vv with respect to tt: aacc=ddt(2at3bt2)=2a6bta_{acc} = \frac{d}{dt}(2at - 3bt^2) = 2a - 6bt.
  3. Condition: We need to find the time tt when the acceleration is zero. Set aacc=0a_{acc} = 0: 2a6bt=02a - 6bt = 0 6bt=2a6bt = 2a t=2a6b=a3bt = \frac{2a}{6b} = \frac{a}{3b}.

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 LINEpositionparticlevariesaccelerationparticle

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