Physics · 2021
JEE Main · 25 July 2021, Shift 2 · Q6
A force F =(40 i +10 j ) N acts on a body of mass 5 kg. If the body starts from rest, its position vector r at time t=10 s, will be:
A force $\displaystyle \overrightarrow{\mathrm{F}}=(40 \hat{i}+10 \hat{j}) \mathrm{N}$ acts on a body of mass $\displaystyle 5$ kg . If the body starts from rest, its position vector $\displaystyle \overrightarrow{\mathrm{r}}$ at time $\displaystyle t=10 \mathrm{~s}$, will be :
Official answer
(1)
$\displaystyle (400 \hat{i}+100 \hat{j}) \mathrm{m}$
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.