The position x of a particle varies with time t as x = at^(2) -bt^(3). For what value of time acceleration is zero?

The position x of a particle varies with time t as x = at^(2) -bt^(3).

1.	The position x of a particle varies with time t as x = at^(2) -bt^(3). For what value of time acceleration is zero?
Answer» `(2a)/(3B)` `a//b` `(a)/(3b)` zero Solution :Here `x=at^(2)-BT^(3)` `v=(dx)/(dt)=2at-3bt^(2)` ACCELERATION =`F=(dv)/(dt)=2a-6bt` Now when f=0 2a-6bt=0 `2a=6bt` or `t=(a)/(3b)`

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The position x of a particle varies with time t as x = at^(2) -bt^(3). For what value of time acceleration is zero?

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