A particle's position (in metres) is (2,3) at t=0, (6,7) at t=2text s , and (13,14) at t=5text s . The average velocity vector from t=0 to t=5text s is:

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NEET Physics QUESTION #7345

Question 1

A particle's position (in metres) is $(2,3)$ at $t=0$, $(6,7)$ at $t=2\text{ s}$, and $(13,14)$ at $t=5\text{ s}$. The average velocity vector from $t=0$ to $t=5\text{ s}$ is:

$\dfrac{1}{5}(13\hat{i}+14\hat{j})$
$\dfrac{7}{3}(\hat{i}+\hat{j})$
$2(\hat{i}+\hat{j})$
$\dfrac{11}{5}(\hat{i}+\hat{j})$✔️

Correct Answer Explanation

Displacement: $\Delta\vec{r} = (13-2)\hat{i} + (14-3)\hat{j} = 11\hat{i}+11\hat{j}$

Average velocity $= \dfrac{\Delta\vec{r}}{\Delta t} = \dfrac{11\hat{i}+11\hat{j}}{5} = \mathbf{\dfrac{11}{5}(\hat{i}+\hat{j})}$

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