AP EAMCET · PHYSICS · Motion In Two Dimensions
A particle moves in the \(x y\)-plane with velocity \(\mathbf{v}=x \hat{\mathbf{i}}+y t \hat{\mathbf{j}}\). At \(t=\frac{x \sqrt{3}}{y}\), the
- A \(\frac{\sqrt{3} y}{2}, \frac{y}{2}\)
- B \(\frac{\sqrt{2} y}{3}, \frac{\sqrt{3} y}{2}\)
- C \(\frac{\sqrt{3} y}{2}, \frac{5 y}{2}\)
- D \(2 \sqrt{3} y, \frac{11 y}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3} y}{2}, \frac{y}{2}\)
Step-by-step Solution
Detailed explanation
Given, velocity of particle is \(\mathbf{v}=x \hat{\mathbf{i}}+y t \hat{\mathbf{j}}\) So, \(|\mathbf{v}|=v=\sqrt{x^2+y^2 t^2}\) Magnitude of tangential acceleration is \[ a_t=\frac{d v}{d t}=\frac{0+2 t y^2}{2 \sqrt{x^2+y^2 t^2}} \] or…
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