MCQEasyJEE Main 2023 · 15 April, Shift 1Velocity & Acceleration

Physics Question from JEE Main 2023 · 15 April, Shift 1

The position of a particle related to time is given by x=(5t2−4t+5) m.x=(5t^2-4t+5)\ \text{m}. The magnitude of velocity of the particle at t=2 st=2\,s will be

  • A

    14 m s−114\ \text{m s}^{-1}

  • B

    16 m s−116\ \text{m s}^{-1}

  • C

    10 m s−110\ \text{m s}^{-1}

  • D

    6 m s−16\ \text{m s}^{-1}

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