Physics · 2021
JEE Main · 17 March 2021, Shift 2 · Q11
The velocity of a particle is v=v_0+ gt + Ft^2. Its position is x=0 at t =0; then its displacement after time ( t =1) is:
The velocity of a particle is $\displaystyle v=v_0+\mathrm{gt}+\mathrm{Ft}^2$. Its position is $\displaystyle x=0$ at $\displaystyle \mathrm{t}=0$; then its displacement after time $\displaystyle (\mathrm{t}=1)$ is :
Official answer
(3)
$\displaystyle v_0+\frac{g}{2}+\frac{F}{3}$
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.