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Step-by-Step Solution
Step 1: Identify the initial velocity
The particle is initially moving eastwards with a velocity of $5 \, \text{m/s}$. We can represent this initial velocity vector as
$ \vec{v}_\text{initial} = 5\,\hat{i}\,\text{m/s} $
(where $ \hat{i} $ is the unit vector in the eastward or $x$-direction).
Step 2: Identify the final velocity
After 10 seconds, the velocity changes to $5 \, \text{m/s}$ northwards. We can represent this final velocity vector as
$ \vec{v}_\text{final} = 5\,\hat{j}\,\text{m/s} $
(where $ \hat{j} $ is the unit vector in the northward or $y$-direction).
Step 3: Compute the change in velocity
The change in velocity $ \Delta \vec{v} $ is given by
$ \Delta \vec{v} = \vec{v}_\text{final} - \vec{v}_\text{initial}
= 5\,\hat{j} - 5\,\hat{i}
= -5\,\hat{i} + 5\,\hat{j} \, \text{(m/s)}. $
Step 4: Calculate the magnitude of the average acceleration
Average acceleration $ \vec{a}_\text{avg} $ is defined as the ratio of change in velocity to the time interval $ \Delta t $:
$ \vec{a}_\text{avg} = \frac{\Delta \vec{v}}{\Delta t}. $
First, find the magnitude of $ \Delta \vec{v} $:
$ |\Delta \vec{v}| = \sqrt{(-5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\,\text{m/s}. $
Since the time interval $ \Delta t $ is 10 seconds, the magnitude of the average acceleration is:
$ |\vec{a}_\text{avg}|
= \frac{|\Delta \vec{v}|}{\Delta t}
= \frac{5\sqrt{2}}{10}
= \frac{1}{\sqrt{2}} \,\text{m/s}^2. $
Step 5: Determine the direction of the average acceleration
The change in velocity vector $ \Delta \vec{v} = -5\,\hat{i} + 5\,\hat{j} $ points diagonally towards the north-west (since the $x$-component is negative and the $y$-component is positive). Therefore, the average acceleration also points towards the north-west.
Final Answer
The average acceleration is
$ \frac{1}{\sqrt{2}} \,\text{m/s}^2 $
towards the north-west.
