A) zero
B) \[\frac{a}{3b}\]
C) \[\frac{2a}{3b}\]
D) \[\frac{a}{b}\]
Correct Answer: B
Solution :
Key Idea: Acceleration and velocity of a particle are given by \[A=\frac{dv}{dt},\] \[v=\frac{dx}{dt}.\] The given equation is \[x=a{{t}^{2}}-b{{t}^{3}}\] Velocity \[v=\frac{dx}{dt}=2at-3b{{t}^{2}}\] Acceleration \[A=\frac{dv}{dt}=2a-6bt\] But A = 0 (given) \[2a-6bt=0\] or \[6bt=2a\] or \[t=\frac{2a}{6b}=\frac{a}{3b}\]You need to login to perform this action.
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