Jeeify
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}

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Sign in to see the solution

Practice more Velocity & Acceleration questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Start free trialMore on Velocity & Acceleration

Related questions