(1997)
Solution:
Given \(x = at^2 - bt^3\). Velocity \(v = \frac{dx}{dt} = 2at - 3bt^2\). Acceleration \(a_c = \frac{dv}{dt} = 2a - 6bt\). For zero acceleration, \(2a - 6bt = 0 \Rightarrow 2a = 6bt \Rightarrow t = \frac{2a}{6b} = \frac{a}{3b}\).
The position (x) of a particle varies with time, (t), as \(x = at^2 - bt^3\). The acceleration will be zero at time (t) equal to:
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