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4-1.Newton's Laws of Motion
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A force $\vec{F}=(40 \hat{i}+10 \hat{j})\, N$ acts on a body of mass $5\, {kg}$. If the body starts from rest, its position vector $\vec{r}$ at time $t=10\, {s}$, will be -
A
$(100 \hat{i}+400 \hat{j}) \,m$
B
$(400 \hat{i}+100 \hat{j})\, m$
C
$(100 \hat{i}+100 \hat{j})\, m$
D
$(400 \hat{i}+400 \hat{j}) \,{m}$
(JEE MAIN-2021)
Solution
$\frac{ d \overrightarrow{ v }}{ dt }=\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=(8 \hat{ i }+2 \hat{ j }) m / s ^{2}$
$\frac{ d \overrightarrow{ r }}{ dt }=\overrightarrow{ v }=(8 t \hat{ i }+2 t \hat{ j }) m / s$
$\overrightarrow{ r }=(8 \hat{ i }+2 \hat{ j }) \frac{ t ^{2}}{2} m$
At $t =10 sec$
$\overrightarrow{ r }=[(8 \hat{ i }+2 \hat{ j }) 50] m$
$\Rightarrow \overrightarrow{ r }=(400 \hat{ i }+100 \hat{ j }) m$
Standard 11
Physics
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