Question

The position of the particle varies with time x=at^2-bt^3.the accelaration of the particle willbe zero at time t

Answer 1

Answer:

As           x=at^2 - bt^3

→      v= dx/dt = 2at-3bt^2

→      a= dv/dt = 2a -6bt

According to question,

when  a=0

 →    2a-6bt = 0

 →    2a= 6bt

 →    t= 2/3b

hence when t=2/3b then the the instantaneos acceleration will be 0.

Answer 2

Answer:

Hii dear

Here is your answer

As           x=at^2 - bt^3

→      v= dx/dt = 2at-3bt^2

→      a= dv/dt = 2a -6bt

According to question,

when  a=0

 →    2a-6bt = 0

 →    2a= 6bt

 →    t= 2/3b

hence when t=2/3b then the the instantaneos acceleration will be 0.

I hope this answer help you

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