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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

The displacement xx of a particle varies with time tt as x=aeαt+beβtx = ae^{-\alpha t} + be^{\beta t}, where a,b,αa, b, \alpha and β\beta are positive constants. The velocity of the particle will:

A

Go on decreasing with time

B

Be independent of α\alpha and β\beta

C

Drop to zero when α=β\alpha = \beta

D

Go on increasing with time

Step-by-Step Solution

  1. Find Velocity (vv): Velocity is the rate of change of displacement (v=dxdtv = \frac{dx}{dt}) . Given x=aeαt+beβtx = ae^{-\alpha t} + be^{\beta t}. Differentiating with respect to tt: v=ddt(aeαt)+ddt(beβt)v = \frac{d}{dt}(ae^{-\alpha t}) + \frac{d}{dt}(be^{\beta t}) v=aαeαt+bβeβtv = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}.
  2. Analyze the Change in Velocity: To determine if the velocity is increasing or decreasing, we look at the acceleration (aacc=dvdta_{acc} = \frac{dv}{dt}) . Differentiating vv with respect to tt: aacc=ddt(aαeαt)+ddt(bβeβt)a_{acc} = \frac{d}{dt}(-a\alpha e^{-\alpha t}) + \frac{d}{dt}(b\beta e^{\beta t}) aacc=aα(α)eαt+bβ(β)eβta_{acc} = -a\alpha(-\alpha)e^{-\alpha t} + b\beta(\beta)e^{\beta t} aacc=aα2eαt+bβ2eβta_{acc} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}.
  3. Conclusion: Since a,b,α,βa, b, \alpha, \beta are positive constants and the exponential function exe^{x} is always positive for any real xx, both terms in the acceleration equation are positive. Therefore, the acceleration aacca_{acc} is always positive (aacc>0a_{acc} > 0). A positive acceleration implies that the velocity is constantly increasing with time.

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 LINEdisplacementparticlevariesaealphabebeta

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