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Find the average acceleration of a particle whose velocity changes from \( 40\hat{i} - 30\hat{j} \) to \( 30\hat{i} \) in 10 seconds.

1. \( 5 \, \text{m/s}^2 \)
2. \( 4 \, \text{m/s}^2 \)
3. \( 3 \, \text{m/s}^2 \)
4. \( 2 \, \text{m/s}^2 \)

Correct answer: \( 4 \, \text{m/s}^2 \)

Solution

The change in velocity is \( \Delta \vec{v} = (30\hat{i} - (40\hat{i} - 30\hat{j})) = -10\hat{i} + 30\hat{j} \). The magnitude of \( \Delta \vec{v} \) is \( \sqrt{(-10)^2 + 30^2} = \sqrt{100 + 900} = 10\sqrt{10} \). Average acceleration is \( \frac{|\Delta \vec{v}|}{\Delta t} = \frac{10\sqrt{10}}{10} = \sqrt{10} \approx 4 \, \text{m/s}^2 \).

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