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NEET PHYSICSEasy

A particle is moving such that its position coordinates (x,y)(x, y) are (2 m,3 m)(2\text{ m}, 3\text{ m}) at time t=0t = 0, (6 m,7 m)(6\text{ m}, 7\text{ m}) at time t=2 st = 2\text{ s} and (13 m,14 m)(13\text{ m}, 14\text{ m}) at time t=5 st = 5\text{ s}. Average velocity vector (vav\mathbf{v}_{av}) from t=0t = 0 to t=5 st = 5\text{ s} is:

A

15(13i^+14j^)\frac{1}{5}(13\hat{i} + 14\hat{j})

B

73(i^+j^)\frac{7}{3}(\hat{i} + \hat{j})

C

2(i^+j^)2(\hat{i} + \hat{j})

D

115(i^+j^)\frac{11}{5}(\hat{i} + \hat{j})

Step-by-Step Solution

  1. Definition: Average velocity (vav\mathbf{v}_{av}) is defined as the ratio of the displacement (Δr\Delta \mathbf{r}) to the corresponding time interval (Δt\Delta t). It depends only on the initial and final positions, not the path taken or intermediate points . vav=r2−r1t2−t1\mathbf{v}_{av} = \frac{\mathbf{r}_2 - \mathbf{r}_1}{t_2 - t_1}
  2. Identify Coordinates:
  • At t1=0t_1 = 0: Position vector r1=2i^+3j^\mathbf{r}_1 = 2\hat{i} + 3\hat{j} meters.
  • At t2=5 st_2 = 5\text{ s}: Position vector r2=13i^+14j^\mathbf{r}_2 = 13\hat{i} + 14\hat{j} meters. (Note: The position at t=2 st=2\text{ s} is irrelevant for the average velocity over the interval 0 to 5 s).
  1. Calculate Displacement (Δr\Delta \mathbf{r}): Δr=r2−r1=(13−2)i^+(14−3)j^=11i^+11j^ m\Delta \mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1 = (13 - 2)\hat{i} + (14 - 3)\hat{j} = 11\hat{i} + 11\hat{j} \text{ m}
  2. Calculate Average Velocity: vav=11i^+11j^5−0=115(i^+j^) m/s\mathbf{v}_{av} = \frac{11\hat{i} + 11\hat{j}}{5 - 0} = \frac{11}{5}(\hat{i} + \hat{j}) \text{ m/s}
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