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Kelvi

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

medium

Motion in a Plane

2014

physics

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

Explanation

To find the average velocity vector $\vec{v}_{av}$ from $t = 0$

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

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

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

to

t = 5s,

we use the formula:

\\ \vec{v}_{av} = \frac{\Delta \vec{r}}{\Delta t} \\

where

\Delta \vec{r}

is the change in position vector and

\Delta t

is the change in time.

\\

Given positions:

\\

• At

t = 0,

position

(x_0, y_0) = (2m, 3m) \\

• At

t = 5s,

position

(x_5, y_5) = (13m, 14m) \\

Calculate the change in position:

\\ \Delta \vec{r} = (x_5 - x_0)\hat{i} + (y_5 - y_0)\hat{j} \\ \Delta \vec{r} = (13m - 2m)\hat{i} + (14m - 3m)\hat{j} \\ \Delta \vec{r} = 11\hat{i} + 11\hat{j} \\

Calculate the change in time:

\\ \Delta t = 5s - 0s = 5s \\

Substitute into the average velocity formula:

\\ \vec{v}_{av} = \frac{11\hat{i} + 11\hat{j}}{5} \\ \vec{v}_{av} = \frac{11}{5} \hat{i} + \frac{11}{5} \hat{j} \\

Thus, the average velocity vector is

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

This corresponds to Option 4.