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JEE Mains · Physics · STD 11 - 3.2 motion in plane
A mosquito is moving with a velocity \(\overrightarrow{ v }=0.5 t ^{2} \hat{ i }+3 t \hat{ j }+9 \hat{ k }\, m / s\) and accelerating in uniform conditions. What will be the direction of mosquito after \(2 \,s\) ?
- A \(\tan ^{-1}\left(\frac{2}{3}\right)\) from \(x-\) axis
- B \(\tan ^{-1}\left(\frac{2}{3}\right)\) from \(y\) -axis
- C \(\tan ^{-1}\left(\frac{5}{2}\right)\) from \(y\) -axis
- D \(\tan ^{-1}\left(\frac{5}{2}\right)\) from \(x\) -axis
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}\left(\frac{2}{3}\right)\) from \(y\) -axis
Step-by-step Solution
Detailed explanation
Given : \(\overrightarrow{ v }=0.5 t ^{2} \hat{ i }+3 t \hat{ j }+9 \hat{ k }\) \(\overrightarrow{ v }_{ at\,t =2}=2 \hat{ i }+6 \hat{ j }+9 \hat{ k }\) \(\therefore\) Angle made by direction of motion of mosquito will be, \(\cos ^{-1} \frac{2}{11}(\) from \(x\) -axis…
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