A force F = (40i + 10j) N acts on a body... | Laws of Motion | JEE Main

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JEE MAIN 2021

25-07-21 S2

Question

A force $\overrightarrow{\mathrm{F}}=(40 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}) \mathrm{N}$ acts on a body of mass 5 kg . If the body starts from rest, its position vector $\overrightarrow{\mathrm{r}}$-at time $\mathrm{t}=10 \mathrm{~s}$, will be .

Select the correct option:

A

$(100 \hat{\mathrm{i}}+400 \hat{\mathrm{j}}) \mathrm{m}$

B

$(100 \hat{\mathrm{i}}+100 \hat{\mathrm{j}}) \mathrm{m}$

C

$(400 \hat{i}+100 \hat{j}) \mathrm{m}$

D

$(400 \hat{\mathrm{i}}+400 \hat{\mathrm{j}}) \mathrm{m}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

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