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JEE Main 2022 (29 Jul, Shift 1) · Velocity, Acceleration & Variable Motion

Velocity, Acceleration & Variable Motion question from JEE Main 2022 (29 Jul, Shift 1)

Q23JEE Main 2022 · 29 JUL · SHIFT 1Medium

Velocity, Acceleration & Variable Motion

If t=x+4t = \sqrt{x} + 4, then (dxdt)t=4\left(\frac{dx}{dt}\right)_{t=4} is:

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Answer: (b)

  1. x=t−4⇒x=(t−4)2\sqrt{x} = t - 4 \Rightarrow x = (t - 4)^2
  2. dxdt=2(t−4)\dfrac{dx}{dt} = 2(t - 4)
  3. At t=4t = 4: dxdt=0\dfrac{dx}{dt} = 0
  4. So the answer is (b) zero.
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