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A particle moves along a straight line such that its acceleration increases at a constant rate of 2 m/s³ (jerk = 2 m/s³). Both the initial acceleration and initial velocity are zero. Find the distance covered by the particle during the 3rd second (i.e., from t = 2 s to t = 3 s).
- 19/3 m
- 12/5 m
- 17/5 m
- 19/4 m
Correct answer: 19/3 m
Solution
Given jerk j = 2 m/s³, a(0) = 0, v(0) = 0. Integrating: a(t) = 2t. v(t) = t². s(t) = t³/3. Since v(t) >= 0 for t >= 0, no reversal. Distance in 3rd second = s(3) - s(2) = 27/3 - 8/3 = 19/3 m.
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