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JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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The position vector of a moving particle at t sec is given by V→=3i^+4t2j^−t3k^.\overrightarrow V = 3\hat i + 4{t^2}\hat j - {t^3}\hat k. Its displacement during an interval from 1s to 3s is.
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Correct option: D
32j^−26k^32\hat j - 26\hat k
Option D (32j^−26k^32\hat j - 26\hat k) is correct because it is the displacement vector calculated by subtracting the initial position vector from the final position vector.

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