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JEE Main 2019 (9 Apr, Shift 2) · Velocity, Acceleration & Variable Motion

Velocity, Acceleration & Variable Motion question from JEE Main 2019 (9 Apr, Shift 2)

Q29JEE Main 2019 · 9 APR · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

The position of a particle as a function of time t, is given by

x(t) = at + bt2 – ct3

where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be:

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Answer: (c)

  1. v=a+2bt−3ct2v = a + 2bt - 3ct^2, acceleration =2b−6ct= 2b - 6ct.
  2. Zero acceleration at t=b3ct = \frac{b}{3c}.
  3. v=a+2bb3c−3cb29c2=a+2b23c−b23cv = a + 2b\frac{b}{3c} - 3c\frac{b^2}{9c^2} = a + \frac{2b^2}{3c} - \frac{b^2}{3c}
  4. So the answer is (c) a+b23ca + \frac{b^2}{3c}.
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