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JEE Main 2020 (9 Jan, Shift 1) · Velocity, Acceleration & Variable Motion

Velocity, Acceleration & Variable Motion question from JEE Main 2020 (9 Jan, Shift 1)

Q28JEE Main 2020 · 9 JAN · SHIFT 1Difficult

Velocity, Acceleration & Variable Motion

The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is __________

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. Differentiate: 2xv=2at+2b⇒xv=at+b2xv = 2at + 2b \Rightarrow xv = at + b
  2. Differentiate again: v2+xac=a⇒ac=a−v2xv^2 + xa_c = a \Rightarrow a_c = \dfrac{a - v^2}{x}
  3. ac=ax2−(at+b)2x3=ac−b2x3a_c = \dfrac{a x^2 - (at + b)^2}{x^3} = \dfrac{ac - b^2}{x^3}
  4. So ac∝x−3a_c \propto x^{-3} and n=3n = 3.
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