A 2lb particle A is acted upon by its weight and the force system: F1={2i+6j-2k}

Question

A 2lb particle A is acted upon by its weight and the force system:
F1={2i+6j-2k}lb, F2={3i-1k}, F3={1i-t^2j-2k}. Determine the distance the particle is from the origin after 3s after it has been released from rest.

I used F=ma for each direction, solved for a, then integrated with respect to t to find v, then plugged everything into s=s0+v0t+(1/2)at^2. Then I took the s vector and solved for the magnitude, getting a final answer of 92.4ft.

The real solution is 745ft. Where did I go wrong and how would I solve this correctly?

Explanation / Answer

weight w = 2 (-k) lb
net force F = F1 + F2 + F3 + w = 6 i + (6 - t2) j - 7 k

mass m = w/g

acceleration a = F/m = g*F/w = 32.2 * [3 i + (3 - t2/2) j - 3.5 k] ft/s2

velocity = a dt = 32.2 * [3t i + (3t - t3/6) j - 3.5 t k] ft/s

displacement s = v dt = 32.2 * [1.5t2 i + (1.5t2 - t4/24) j - 1.75t2 k] ft

plug t = 3

s = 32.2 * [1.5*9 i + (1.5*9 - 81/24) j - 1.75*9 k] ft

distance = 32.2 * [(1.5*9)2 + (1.5*9 - 81/24)2 + (1.75*9)2] = 743 ft

plug

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