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4-1.Newton's Laws of Motion
medium
A force on an object of mass $100\, g$ is $(10 \hat{i}+5 \hat{j}) N$. The position of that object at $t=$ $2 s$ is $(a \hat{i}+b \hat{j}) m$ after starting from rest. The value of $\frac{a}{b}$ will be..........
A
$2$
B
$99$
C
$20$
D
$60$
(JEE MAIN-2022)
Solution
$\overrightarrow{ F }=10 \hat{ i }+5 \hat{ j }$
$m =100 \,g =0.1 \,kg$
$\vec{a}=\frac{\overrightarrow{ F }}{ m }=100 \hat{ i }+50 \hat{ j }$
$\overrightarrow{ S }=\overrightarrow{ u t}+\frac{1}{2} \overrightarrow{ a }{ }^{2}=\frac{1}{2} \overrightarrow{ a } t ^{2}($ as $\overrightarrow{ u }=0)$
$=\frac{1}{2}(100 \hat{ i }+50 \hat{ j }) 2^{2}$
$=200 \hat{i}+100 \hat{j}$
$=a \hat{i}+b \hat{j}$
$a=200, b=100$
$\therefore \frac{a}{b}=2$
Standard 11
Physics
