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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 -

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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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