TS EAMCET · Physics · Motion In One Dimension
A particle initially at origin starts moving in xy-plane has velocity component \(\overrightarrow{\mathrm{V}}=(6+2 t) \hat{i}+(4+2 \sqrt{3} t) \hat{j} \mathrm{~m} / \mathrm{s}\).
Acceleration of the particle in \(\mathrm{m} / \mathrm{s}^2\) is \([\mathrm{x}, \mathrm{y}\) are measured in metres, \(t\) in seconds, respectively]
- A \((6+2 t) \hat{i}+(4+2 \sqrt{3} t) \hat{j}\)
- B \((6+2 t) \hat{i}+2 \sqrt{3} \hat{j}\)
- C \(2 \hat{i}+2 \sqrt{3} \hat{j}\)
- D \(2 \hat{i}+2 \sqrt{3} \hat{k}\)
Answer & Solution
Correct Answer
(C) \(2 \hat{i}+2 \sqrt{3} \hat{j}\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & a_x=\frac{d V_x}{d t}=\frac{d}{d t}(6+2 t)=2 \\ & a_y=\frac{d V_y}{d t}=\frac{d}{d t}(4+2 \sqrt{3} t)=2 \sqrt{3} \\ & \text { So, } \vec{a}=a_x \hat{i}+a_y \hat{j} \end{aligned} \] \(=2 \hat{i}+2 \sqrt{3} \hat{j}\)
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