Physics · 2023
JEE Main · 15 April 2023, Shift 1 · Q38
The position of a particle related to time is given by x=(5 t^2-4 t+5) m. The magnitude of velocity of the particle at t=2 s will be:
The position of a particle related to time is given by $\displaystyle x=\left(5 t^2-4 t+5\right) \mathrm{m}$. The magnitude of velocity of the particle at $\displaystyle t=2 s$ will be:
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 16 \mathrm{~ms}^{-1}$
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JEE Main 2023 Physics question, with the answer from NTA’s final answer key. Where our answers come from.