TS EAMCET · Physics · Motion In One Dimension
Two persons \(A\) and \(B\) are located in \(X-Y\) plane at the points \((0,0)\) and \((0,10)\) respectively. (The distances are measured in MKS unit). At a time \(t=0\), they start moving simultaneously with velocities \(\quad \overrightarrow{\mathbf{v}}_A=2 \mathbf{j} \mathrm{ms}^{-1} \quad\) and \(\quad \overrightarrow{\mathbf{v}}_B=2 \hat{\mathbf{i}} \mathrm{ms}^{-1}\) respectively. The time after which \(A\) and \(B\) are at their closest distance is
- A 2.5 s
- B 4 s
- C 1 s
- D \(\frac{10}{\sqrt{2}} \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) 4 s
Step-by-step Solution
Detailed explanation
Let after the time \((t)\) the position of \(A\) is \(\left(0, v_A t\right)\) and position of \(B=\left(v_B t, 10\right)\). Distance between them…
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