JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \((400 \hat{i}+400 \hat{j}) \,{m}\)
Answer & Solution
Correct Answer
(B) \((400 \hat{i}+100 \hat{j})\, m\)
Step-by-step Solution
Detailed explanation
\(\frac{ d \overrightarrow{ v }}{ dt }=\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=(8 \hat{ i }+2 \hat{ j }) m / s ^{2}\) \(\frac{ d \overrightarrow{ r }}{ dt }=\overrightarrow{ v }=(8 t \hat{ i }+2 t \hat{ j }) m / s\)…
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