Physicshard · Past Paper
If the velocity of a particle is given by v = At + Bt^2, where A and B are constants, then the distance traveled by it between 1s and 2s is:
A3A/2 + 7B/3
BA/2 + B/3
C3A/2 + B/3
DA + B
✓ Correct Answer: A — 3A/2 + 7B/3
Distance = integral from 1 to 2 of (At + Bt^2)dt = [At^2/2 + Bt^3/3] from 1 to 2 = (2A + 8B/3) - (A/2 + B/3) = 1.5A + 7/3B.
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