The displacement of a particle varies with time t as: s=αt2−βt3. Determine the time (t) at which acceleration of the particle becomes 0.
A
α3β
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B
αβ
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C
3βα
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D
2α3β
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Solution
The correct option is Aα3β Given, s=αt2−βt3 Differentiating s w.r.t t we get, Velocity, v=dsdt=2αt−3βt2 Differentiating v w.r.t t we get, Acceleration, a=dvdt=2α−6βt As a=0, we get 2α−6βt=0⇒t=α3β