JEE Main 202524 Jan, Evening shiftPhysicsEasyKinematics177× since 2002The position vector of a moving body at any instant of time is given as r⃗=(5t2i^−5tj^)\vec r = (5t^2\hat i - 5t\hat j)r=(5t2i^−5tj^) m. The magnitude and direction of velocity at t=2t = 2t=2 s is,AOption A. 5155\sqrt{15}515 m/s, making an angle of tan−14\tan^{-1}4tan−14 with −-−ve Y axisBOption B. 5155\sqrt{15}515 m/s, making an angle of tan−14\tan^{-1}4tan−14 with +++ve X axisCOption C. 5175\sqrt{17}517 m/s, making an angle of tan−14\tan^{-1}4tan−14 with +++ve X axisDOption D. 5175\sqrt{17}517 m/s, making an angle of tan−14\tan^{-1}4tan−14 with −-−ve Y axisCheckSkip · show solution
