Average velocity from position coordinates – Rankers Physics

Average Speed and Velocity: Practice Problem & Solution

A particle is moving such that its position coordinates (x, y) are: \((2\text{ m}, 3\text{ m})\text{ at time } t = 0,\) \((6\text{ m}, 7\text{ m})\text{ at time } t = 2\text{ s}\) and \((13\text{ m}, 14\text{ m})\text{ at time } t = 5\text{ s}\). Average velocity \((\vec{V}_{av})\text{ from } t = 0\text { to } t = 5\text{ s}\) is: (2014)
\(\frac{1}{5}(13\hat{i}+14\hat{j})\)
\(\frac{7}{3}(\hat{i}+\hat{j})\)
\(2(\hat{i}+\hat{j})\)
\(\frac{11}{5}(\hat{i}+\hat{j})\)

Solution Explained:

To solve this problem, we apply the core principles of Average Speed and Velocity. Understanding the underlying formula is key to arriving at the correct answer below:

Average velocity is \(\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\vec{r}_f - \vec{r}_i}{t_f - t_i}\). Initial position at \(t_i = 0\text{ s}\) is \(\vec{r}_i = 2\hat{i} + 3\hat{j}\). Final position at \(t_f = 5\text{ s}\) is \(\vec{r}_f = 13\hat{i} + 14\hat{j}\). So, \(vec{V}_{av} = \frac{(13\hat{i} + 14\hat{j}) - (2\hat{i} + 3\hat{j})}{5 - 0} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j})\) m/s.

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