JEE Main 202315 Apr, Morning shiftPhysicsMediumKinematics177× since 2002The position of a particle related to time is given by x=(5t2−4t+5)mx=\left(5 t^{2}-4 t+5\right) \mathrm{m}x=(5t2−4t+5)m. The magnitude of velocity of the particle at t=2st=2 st=2s will be :AOption A. 14 ms−114 \mathrm{~ms}^{-1}14 ms−1BOption B. 16 ms−116 \mathrm{~ms}^{-1}16 ms−1COption C. 10 ms−110 \mathrm{~ms}^{-1}10 ms−1DOption D. 06 ms−106 \mathrm{~ms}^{-1}06 ms−1CheckSkip · show solution
