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MCQEasyJEE Main 2024 · 29 January, Shift 2Velocity & Acceleration

Physics Question from JEE Main 2024 · 29 January, Shift 2

A particle is moving in a straight line. The variation of position xx as a function of time tt is given as x=t3−6t2+20t+15 mx = t^3 - 6t^2 + 20t + 15 \, \text{m}. The velocity of the body when its acceleration becomes zero is:

  • A

    4 m/s4 \, \text{m/s}

  • B

    8 m/s8 \, \text{m/s}

  • C

    10 m/s10 \, \text{m/s}

  • D

    6 m/s6 \, \text{m/s}

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