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Given: F = 6ti + 4tj, Fx = 6t, Fy = 4t, ax = 6t/3 = 2t, ay = 4t/3, dx/dt = 2t^2, ∫Vy dVy = ∫2t^2 dt

1. (a) vx = 2/3·3^3 = 2/3·27 = 18
2. (b) dVy/dt = 4/3·∫Vy dt
3. (c) ∫Vy dVy = 4/3·∫Vy dt
4. (d) ∫Vy dVy = 4/3·∫Vy dt

Correct answer: (a) vx = 2/3·3^3 = 2/3·27 = 18

Solution

The correct answer is (a). The velocity in the x-direction, vx, is calculated by integrating the acceleration ax = 2t with respect to time, which gives vx = (2/3)t^3. Substituting t = 3, we get vx = 2/3·27 = 18.

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