Q128 / 150 · MHT CET 2024 (16 May Shift 1) · Physics

The co-ordinates of a moving particle at any time ' t ' are given by $x=\alpha \mathrm{t}^3$ and $y=\beta \mathrm{t}^3$ where $\alpha$ and $\beta$ are constants. The speed of the particle at time ' $t$ ' is given by

  • A
    $t \sqrt{\alpha^2+\beta^2}$
  • B
    $3 \mathrm{t} \sqrt{\alpha^2+\beta^2}$
  • C
    $t^2 \sqrt{\alpha^2+\beta^2}$
  • Correct answer
    D
    $3 \mathrm{t}^2 \sqrt{\alpha^2+\beta^2}$

Solution

$\begin{aligned} & X=\alpha t^3 \hat{i}+\beta t^3 \hat{j} \\ & \begin{aligned} V=\frac{d X}{d t}=3 \alpha t^2 \hat{i} & +3 \beta t^2 \hat{j} \\ \text { Magnitude of } V & =\sqrt{\left(3 \alpha t^2\right)^2+\left(3 \beta t^2\right)^2} \\ & =\sqrt{9 \alpha^2 t^4+9 \beta^2 t^4} \\ & =\sqrt{9 t^4\left(\alpha^2+\beta^2\right)} \\ & =3 t^2 \sqrt{\alpha^2+\beta^2}\end{aligned}\end{aligned}$

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

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