Q53 / 150 · MHT CET 2025 (19 Apr Shift 2) · Mathematics

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

  • A
    $\mathrm{b}-\mathrm{c} \quad$ unit/seconds ${ }^2$
  • B
    $\mathrm{b}+\mathrm{c} \quad$ unit/seconds ${ }^2$
  • C
    $2 \mathrm{~b}-2 \mathrm{c} \quad$ unit $/$ seconds $^2$
  • Correct answer
    D
    $2 \sqrt{b^2+c^2}$ unit/seconds ${ }^2$

Solution

Differentiating the position equations yields velocity components:

$v_x = \frac{dx}{dt} = b - 2ct$
$v_y = \frac{dy}{dt} = a + 2bt$

Differentiating again gives the acceleration components:

$a_x = \frac{dv_x}{dt} = -2c$
$a_y = \frac{dv_y}{dt} = 2b$

The magnitude of the resultant acceleration is therefore:

$A = \sqrt{(-2c)^2 + (2b)^2} = \sqrt{4c^2 + 4b^2} = \sqrt{4(b^2 + c^2)} = 2\sqrt{b^2 + c^2}$

Final answer: $\boxed{\text{D}}$

Hence, the correct answer is option: 'D'.

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