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NEET PHYSICSEasy

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. The average velocity vector v_avg from t = 0 to t = 5 s is:

A

1/5 (13î + 14ĵ)

B

7/3 (î + ĵ)

C

2 (î + ĵ)

D

11/5 (î + ĵ)

Step-by-Step Solution

Average velocity is defined as the displacement divided by the corresponding time interval . v⃗avg=Δr⃗Δt=r⃗f−r⃗itf−ti\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\vec{r}_f - \vec{r}_i}{t_f - t_i}

  1. Identify Initial and Final Positions: At t1=0t_1 = 0 s, position vector r⃗1=2i^+3j^\vec{r}_1 = 2\hat{i} + 3\hat{j} m. At t2=5t_2 = 5 s, position vector r⃗2=13i^+14j^\vec{r}_2 = 13\hat{i} + 14\hat{j} m.

  2. Calculate Displacement (Δr⃗\Delta \vec{r}): Δr⃗=r⃗2−r⃗1=(13i^+14j^)−(2i^+3j^)\Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (13\hat{i} + 14\hat{j}) - (2\hat{i} + 3\hat{j}) Δr⃗=(13−2)i^+(14−3)j^=11i^+11j^ m\Delta \vec{r} = (13 - 2)\hat{i} + (14 - 3)\hat{j} = 11\hat{i} + 11\hat{j} \text{ m}

  3. Calculate Time Interval (Δt\Delta t): Δt=5−0=5 s\Delta t = 5 - 0 = 5 \text{ s}

  4. Calculate Average Velocity: v⃗avg=11i^+11j^5=115(i^+j^) m/s\vec{v}_{avg} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j}) \text{ m/s}

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