Laws of Motion

JEE Main level, single correct, medium

34 of 210

Question 34 of 210

A force F⃗=(40i^+10j^)N\vec{F}=(40\hat{i}+10\hat{j})N acts on a body of mass 5kg5kg. If the body starts from rest, its position vector r⃗\vec{r} at time t=10st=10s will be

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Answer: (C) (400i^+100j^)m(400\hat{i}+100\hat{j})m

dv⃗dt=a⃗=F⃗m=(8i^+2j^)ms−2\dfrac{d\vec{v}}{dt}=\vec{a}=\dfrac{\vec{F}}{m}=(8\hat{i}+2\hat{j})m{s}^{-2}
dr⃗dt=v⃗=(8ti^+2tj^)ms−1\dfrac{d\vec{r}}{dt}=\vec{v}=(8t\hat{i}+2t\hat{j})m{s}^{-1}
r⃗=(i^+2j^)t22m\vec{r}=(\hat{i}+2\hat{j})\dfrac{{t}^{2}}{2}m
At t=10sec⁡t=10\sec
r⃗=[(8i^+2j^)50]m\vec{r}=[(8\hat{i}+2\hat{j})50]m
⇒r⃗=(400i^+100j^)m\Rightarrow \vec{r}=(400\hat{i}+100\hat{j})m

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