Question:
A particle moves from the point \( (2.0\hat{i} + 4.0\hat{j}) \) m, at t = 0, with an initial velocity \( (5.0\hat{i} + 4.0\hat{j}) \) \( \mathrm{ms^{-1}} \). It is acted upon by a constant force which produces a constant acceleration \( (4.0\hat{i} + 4.0\hat{j}) \) \( \mathrm{ms^{-2}} \). What is the distance of the particle from the origin at time 2 s?
A particle moves from the point \( (2.0\hat{i} + 4.0\hat{j}) \) m, at t = 0, with an initial velocity \( (5.0\hat{i} + 4.0\hat{j}) \) \( \mathrm{ms^{-1}} \). It is acted upon by a constant force which produces a constant acceleration \( (4.0\hat{i} + 4.0\hat{j}) \) \( \mathrm{ms^{-2}} \). What is the distance of the particle from the origin at time 2 s?
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Use \( \vec{r} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2 \) component-wise and find the magnitude.
Updated On: Nov 5, 2026
- 15 m
- \( 20\sqrt{2} \) m
- \( 10\sqrt{2} \) m
- 5 m
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The Correct Option is B
Solution and Explanation
\( \vec{r} = \vec{r}_0 + \vec{u}t + \frac12\vec{a}t^2 \) with t = 2 s.
x: \( 2 + 5(2) + \frac12(4)(4) = 2 + 10 + 8 = 20 \) m.
y: \( 4 + 4(2) + \frac12(4)(4) = 4 + 8 + 8 = 20 \) m.
Distance from origin \( = \sqrt{20^2 + 20^2} = 20\sqrt{2} \) m.
Answer: (2) \( 20\sqrt{2} \) m
x: \( 2 + 5(2) + \frac12(4)(4) = 2 + 10 + 8 = 20 \) m.
y: \( 4 + 4(2) + \frac12(4)(4) = 4 + 8 + 8 = 20 \) m.
Distance from origin \( = \sqrt{20^2 + 20^2} = 20\sqrt{2} \) m.
Answer: (2) \( 20\sqrt{2} \) m
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