JEE Mains · Physics · STD 11 - 3.2 motion in plane
A butterfly is flying with a velocity \(4 \sqrt{2} \,{m} / {s}\) in North-East direction. Wind is slowly blowing at \(1\) \({m} / {s}\) from North to South. The resultant displacement of the butterfly in \(3\, seconds\) is \(....\,{m}.\)
- A \(15\)
- B \(122\)
- C \(3\)
- D \(20\)
Answer & Solution
Correct Answer
(A) \(15\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{{V}}_{{BW}}=4 \sqrt{2} \cos 45 \hat{{i}}+4 \sqrt{2} \sin 45 \hat{{j}}\) \(=4 \hat{{i}}+4 \hat{{j}}\) \(\overrightarrow{{V}}_{{w}}=-\hat{{j}}\) \(\overrightarrow{{V}}_{{B}}=\overrightarrow{{V}}_{{Bw}}+\overrightarrow{{V}}_{{T}}=4 \hat{{i}}+3 \hat{{j}}\)…
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