Q. A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s. Average velocity vector $(\vec{V}_{av})$ from t = 0 to t = 5 s is :
A) $\frac{1}{5}(13\hat{i} + 14\hat{j})$
B) $\frac{7}{3}(\hat{i} + \hat{j})$
C) $2(\hat{i} + \hat{j})$
D) $\frac{11}{5}(\hat{i} + \hat{j})$
Explanation
[Solution] Displacement = $(13-2)\hat{i} + (14-3)\hat{j} = 11(\hat{i} + \hat{j})$ in 5 s, so $\vec{V}_{av} = \frac{11}{5}(\hat{i} + \hat{j})$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.