A force F = (40i + 10j) N acts on a body of mass 5 kg. If the body starts from rest, its position vector r at time t = 10 s, will be: (a) (100i + 400j) m (b) (100i + 100j) m (c) (400i + 100j) m (d) (400i + 400j) m

Answer: (c). a = (40i + 10j)/(5) = 8i + 2j m/s² r = 1/2at² = 50(8i + 2j) = 400i + 100j m So the answer is (c).

A force F = (40i + 10j) N acts on a body of mass 5 kg. If the body… | JEE Main 2021 (25 Jul, Shift 2)

Q22JEE Main 2021 · 25 JUL · SHIFT 2Medium

Inertia, First & Second Laws

A force $\vec{F} = (40\hat{i} + 10\hat{j})\text{ N}$ 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: (c)

$\vec{a} = \frac{40\hat{i} + 10\hat{j}}{5} = 8\hat{i} + 2\hat{j}\text{ m/s}^2$
$\vec{r} = \frac{1}{2}\vec{a}t^2 = 50(8\hat{i} + 2\hat{j}) = 400\hat{i} + 100\hat{j}\text{ m}$
So the answer is (c).

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