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Step-by-Step Solution
Step 1: Identify the initial and final velocities
• The body initially has a velocity of $30\,\text{m/s}$ directed towards the east. We can represent this as a vector:
$$ \vec{v_i} = 30\,\hat{i}\,\text{m/s} $$
• After $10$ seconds, the velocity becomes $40\,\text{m/s}$ directed towards the north. We can represent this as:
$$ \vec{v_f} = 40\,\hat{j}\,\text{m/s} $$
Step 2: Calculate the change in velocity
The change in velocity (vector) is:
$$ \Delta \vec{v} = \vec{v_f} - \vec{v_i} = 40\,\hat{j} - 30\,\hat{i} \text{ (m/s)}. $$
Step 3: Find the magnitude of the change in velocity
The components of $\Delta \vec{v}$ are:
x-component: $-30\,\text{m/s}$ (since it changes from $30\,\hat{i}$ to $0$ in the x-direction).
y-component: $+40\,\text{m/s}$.
Thus, its magnitude is:
$$ |\Delta \vec{v}| = \sqrt{(-30)^2 + (40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\,\text{m/s}. $$
Step 4: Find the time interval
The time interval $ \Delta t $ given is $10\,\text{s}.$
Step 5: Compute the average acceleration
Average acceleration is defined as the change in velocity divided by the time interval:
$$ \vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t}. $$
Since the magnitude of the change in velocity is $50\,\text{m/s}$ over $10\,\text{s}$, the magnitude of the average acceleration is:
$$ | \vec{a}_{\text{avg}} | = \frac{50\,\text{m/s}}{10\,\text{s}} = 5\,\text{m/s}^2. $$
Step 6: Final Answer
Therefore, the average acceleration of the body is
$$5\,\text{m/s}^2.$$
