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JEE Main 2011 (AIEEE) · Velocity, Acceleration & Variable Motion

Velocity, Acceleration & Variable Motion question from JEE Main 2011 (AIEEE)

Q37JEE Main 2011 · AIEEEMedium

Velocity, Acceleration & Variable Motion

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:

dvdt=−2.5v\frac{dv}{dt} = -2.5\sqrt{v} where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

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Answer: (a)

  1. dvv=−2.5 dt\dfrac{dv}{\sqrt{v}} = -2.5\,dt
  2. ∫6.250v−1/2dv=−2.5 t⇒−26.25=−2.5 t\int_{6.25}^{0} v^{-1/2}dv = -2.5\,t \Rightarrow -2\sqrt{6.25} = -2.5\,t
  3. t=2×2.52.5=2 st = \dfrac{2 \times 2.5}{2.5} = 2\text{ s}
  4. So the answer is (a) 2 s.
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