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. Average velocity vector $ ( \vec{ v_{av}})$ from t = 0 to t = 5 s is .

Question

A ball is thrown vertically upwards with an initial velocity of $ 20 \, \text{m/s} $. How high will the ball rise? (Take $ g = 10 \, \text{m/s}^2 $)

Answer 1

To find the average velocity vector $ (\vec{v_{av}})$ for the particle moving from time $ t = 0 \, \text{s} $ to $ t = 5 \, \text{s} $, we will use the formula for average velocity:

$$ \vec{v_{av}} = \frac{\Delta \vec{r}}{\Delta t} $$

where $ \Delta \vec{r} $ is the change in position vector, and $ \Delta t $ is the change in time.

The initial position vector at $ t = 0 \, \text{s} $ is $ \vec{r_1} = 2 \widehat{i} + 3 \widehat{j} $.
The final position vector at $ t = 5 \, \text{s} $ is $ \vec{r_2} = 13 \widehat{i} + 14 \widehat{j} $.
Calculate the change in the position vector:

$$ \Delta \vec{r} = \vec{r_2} - \vec{r_1} = (13 \widehat{i} + 14 \widehat{j}) - (2 \widehat{i} + 3 \widehat{j}) $$

Simplifying, we get:

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

Calculate the time interval:

$$ \Delta t = 5 \, \text{s} - 0 \, \text{s} = 5 \, \text{s} $$

Now, compute the average velocity vector:

$$ \vec{v_{av}} = \frac{\Delta \vec{r}}{\Delta t} = \frac{11 \widehat{i} + 11 \widehat{j}}{5} = \frac{11}{5} \widehat{i} + \frac{11}{5} \widehat{j} $$

Thus, the average velocity vector from $ t = 0 \, \text{s} $ to $ t = 5 \, \text{s} $ is $ \frac{11}{5} ( \widehat{i} + \widehat{j}) $.

Hence, the correct answer is: $$ \frac{11}{5} ( \widehat{i} + \widehat{j}) $$.

Answer 2