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MHT CET202519 Apr 2025Evening ShiftMathematicsApplication of DerivativesActual

The position of a point in time t is given by x= a + bt - ct ^2, y = at + bt ^2 . It's resultant acceleration at time t in seconds is given by

Options

  1. Ab - c unit/seconds ^2
  2. Bb + c unit/seconds ^2
  3. C2 ~b -2 c unit / seconds ^2
  4. D2 b^2+c^2 unit/seconds ^2

Correct answer

D. 2 b^2+c^2 unit/seconds ^2

Step-by-step solution

Differentiating the position equations yields velocity components: v_x = dx dt = b - 2ct v_y = dy dt = a + 2bt Differentiating again gives the acceleration components: a_x = dv_x dt = -2c a_y = dv_y dt = 2b The magnitude of the resultant acceleration is therefore: A = (-2c)^2 + (2b)^2 = 4c^2 + 4b^2 = 4(b^2 + c^2) = 2 b^2 + c^2 Final answer: D

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