Verified Solution • 2025–26 Session moderate

QJEEM25Pha158: JEE Main 2025 Physics - Class 11 JEEM25Pha158 Solution: The position vector of a moving body at any instant of time is…

The position vector of a moving body at any instant of time is given as r→=(5t2i^−5tj^)m\overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m} . The magnitude and direction of velocity at t=2st=2 s is,

Multiple Choice Options

  • A 517 m/s5 \sqrt{17} \mathrm{~m} / \mathrm{s} , making an angle of tan⁡−14\tan ^{-1} 4 with - ve Y axis
  • B 515 m/s5 \sqrt{15} \mathrm{~m} / \mathrm{s} , making an angle of tan⁡−14\tan ^{-1} 4 with + ve XX axis
  • C 517 m/s5 \sqrt{17} \mathrm{~m} / \mathrm{s} , making an angle of tan⁡−14\tan ^{-1} 4 with + ve XX axis
  • D 515 m/s5 \sqrt{15} \mathrm{~m} / \mathrm{s} , making an angle of tan⁡−14\tan ^{-1} 4 with −- ve YY axis

Step-by-Step Verified Solution

r→=5t2i^−5t^v→=10ti^−5j^v→=20i^−5j^ at t=2sec

JEE Main 2025 (Online) 24th January Evening Shift Physics - Motion in a Plane Question 1 English Explanation

tan⁡θ=205=4θ=tan−1⁡4 From-veY-axis 

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