The distance-time graph of a moving particle is given by `x = 4t -6t^(2)`. What is the positive maximum speed?

Explanation

Velocity of the particle v = `(dx)/(dt) =` 4-12t
Acceleration of the particle, a = `(d^(2)x)/(dt^(2))=-12 m//s^(2)`
So, the particle is moving with retardation at t = 0 the velocity is maximum. In this case
(i) the maximum positive velocity of the particle = 4-12`xx0 = 4 m//s`