Solution:
Differentiating position equations twice yields accelerations: \(a_x = 48\text{ m/s}^2\) and \(a_y = 18t\text{ m/s}^2\). At \(t = 2\text{ s}\), \(a_y = 36\text{ m/s}^2\). Total acceleration \(a = \sqrt{a_x^2 + a_y^2} = 60\text{ m/s}^2\). The net force is \(F = ma = 3.0 \times 60 = 180\text{ N}\).
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