Question

a position of a particle with time
as x=at2-bt3
the acceleration of particle zero at time t which will be equal to options:-
$\frac{2a}{3b}$
$\frac{a}{b}$
$\frac{a}{3b}$
$0$


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Answer 1

Answer:

3

Explanation:

differentiation of X

would be velocity = 2at-3bt^2

differentiation of velocity

would be acceleration = 2a-6bt

putting acceleration =0

t = a/3b