Motion in a Plane - Result Question 4
4. A particle is moving such that its position coordinate $(x, y)$ 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 $( \vec{V} _{a v})$ from $t=0$ to $t=5 s$ is :
[2014]
(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) $\frac{11}{5}(\hat{i}+\hat{j})$
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Answer:
Correct Answer: 4. (d)
Solution:
(d) $ \vec{v} _{a v}=\frac{\Delta \vec{r}(\text{ displacement })}{\Delta t(\text{ time taken })}$
$=\frac{(13-2) \hat{i}+(14-3) \hat{j}}{5-0}=\frac{11}{5}(\hat{i}+\hat{j})$ When a point have coordinate $(x, y)$ then its position vector $=x \hat{i}+y \hat{j}$
When a particle moves from point $(x_1, y_1)$ to $(x_2, y_2)$ then its displacement vector
$\vec{r}=(x_2-x_1) \hat{i}+(y_2-y_1) \hat{j}$
