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A particle starts with an initial velocity (i + j) m/s and after 2 seconds its velocity becomes (3i - 5j) m/s. Assuming constant acceleration, what is the magnitude of the displacement of the particle in 2 seconds?

1. 2*sqrt(2) m
2. 3*sqrt(2) m
3. 4*sqrt(2) m
4. sqrt(2) m

Correct answer: 4*sqrt(2) m

Solution

With constant acceleration, displacement = (u + v)/2 * t = ((1+3)/2 * i + (1-5)/2 * j) * 2 = (2i - 2j) * 2 = 4i - 4j m. Magnitude = sqrt(16 + 16) = sqrt(32) = 4*sqrt(2) m.

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