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EEJ MAIN Physics QUESTION #7093
Question 1
A particle starts at $(2, 4)$ at $t=0$ with velocity $(5\hat{i}+4\hat{j})$. Constant acceleration is $(4\hat{i}+4\hat{j})$. Find its distance from the origin at $t=2\text{ s}$.
  • 15 m
  • $20\sqrt{2}\text{ m}$✔️
  • 5 m
  • $10\sqrt{2}\text{ m}$
Correct Answer Explanation

Using $\vec{s} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$:

  • $x = 2 + 5(2) + \frac{1}{2}(4)(2^2) = 2 + 10 + 8 = 20$
  • $y = 4 + 4(2) + \frac{1}{2}(4)(2^2) = 4 + 8 + 8 = 20$
  • Distance from origin $D = \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m}$.