MHT CET · Physics · Motion In One Dimension
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 \(t^2 \sqrt{\alpha^2+\beta^2}\)
- D \(3 \mathrm{t}^2 \sqrt{\alpha^2+\beta^2}\)
Answer & Solution
Correct Answer
(D) \(3 \mathrm{t}^2 \sqrt{\alpha^2+\beta^2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & X=\alpha t^3 \hat{i}+\beta t^3 \hat{j} \\ & \begin{aligned} V=\frac{d X}{d t}=3 \alpha t^2 \hat{i} & +3 \beta t^2 \hat{j} \\ \text { Magnitude of } V & =\sqrt{\left(3 \alpha t^2\right)^2+\left(3 \beta t^2\right)^2} \\ & =\sqrt{9 \alpha^2 t^4+9 \beta^2 t^4} \\ & =\sqrt{9 t^4\left(\alpha^2+\beta^2\right)} \\ & =3 t^2 \sqrt{\alpha^2+\beta^2}\end{aligned}\end{aligned}\)
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