JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2021
- Medium
- MCQ
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 -
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \((400 \hat{i}+100 \hat{j})\, m\)
Step-by-step Solution
Detailed explanation
\(\frac{ d \overrightarrow{ v }}{ dt }=\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=(8 \hat{ i }+2 \hat{ j }) m / s ^{2}\)
\(\frac{ d \overrightarrow{ r }}{ dt }=\overrightarrow{ v }=(8 t \hat{ i }+2 t \hat{ j }) m / s\)
\(\overrightarrow{ r }=(8 \hat{ i }+2 \hat{ j }) \frac{ t ^{2}}{2} m\)
At \(t =10 sec\)
\(\overrightarrow{ r }=[(8 \hat{ i }+2 \hat{ j }) 50] m\)
\(\Rightarrow \overrightarrow{ r }=(400 \hat{ i }+100 \hat{ j }) m\)
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