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Motion in A Straight Line question

2011 · Shift 0 · Q78
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Motion in A Straight Line question

2011 · Shift 0 · Q78

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by : dvdt=−2.5v{{dv} \over {dt}} = - 2.5\sqrt vdtdv​=−2.5v​ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :
  1. A
    2 s
  2. B
    4 s
  3. C
    8 s
  4. D
    1 s
View written solutionFree

Correct answer: A

  1. Given differential equation

The retardation is given by dvdt=−2.5v\frac{dv}{dt}=-2.5\sqrt{v}dtdv​=−2.5v​ with initial speed v0=6.25 m/sv_0=6.25\ \text{m/s}v0​=6.25 m/s and final speed at rest v=0.v=0.v=0.

  1. Separate variables

dvv=−2.5 dt\frac{dv}{\sqrt{v}}=-2.5\,dtv​dv​=−2.5dt

  1. Integrate both sides

From v=6.25v=6.25v=6.25 at t=0t=0t=0 to v=0v=0v=0 at t=Tt=Tt=T:

∫6.250v−1/2 dv=∫0T−2.5 dt\int_{6.25}^{0} v^{-1/2}\,dv = \int_0^T -2.5\,dt∫6.250​v−1/2dv=∫0T​−2.5dt

Since ∫v−1/2 dv=2v,\int v^{-1/2}\,dv = 2\sqrt{v},∫v−1/2dv=2v​, we get

[2v]6.250=−2.5T\left[2\sqrt{v}\right]_{6.25}^{0} = -2.5T[2v​]6.250​=−2.5T

  1. Evaluate limits

2(0−6.25)=−2.5T2(\sqrt{0}-\sqrt{6.25})=-2.5T2(0​−6.25​)=−2.5T

2(0−2.5)=−2.5T2(0-2.5)=-2.5T2(0−2.5)=−2.5T

−5=−2.5T-5=-2.5T−5=−2.5T

T=2 sT=2\ \text{s}T=2 s

  1. Check options
  • A: 2 2\,2s ✅
  • B: 4 4\,4s ❌
  • C: 8 8\,8s ❌
  • D: 1 1\,1s ❌

Therefore, the correct option is A.

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