Question

The position x of a particle varies with time to as x= at square - bt cube. The acceleration of the particle will be zero at time t equal to
1) a/b 2) 2a/3b 3) a/3b 4) zero

Answer 1

differentiate the function two times with respect to t . you will get acceleration and equate it to O as given in question .
after that you will find your ans t = a / 3b

Answer 2

x = at² - bt³

acceleration means change in velocity per unit time . e.g a = dv/dt .
and we also know, velocity is change in displacement per unit time .e.g v = dx/dt

hence, acceleration = d²x/dt²

now, differentiate x = at² - bt³ with respect to t

dx/dt = 2at - 3bt²

again, differentiate with respect to t
d²x/dt² = 2a - 6bt

according to above we know, d²x/dt² is acceleration so,
acceleration = 2a - 6bt
but question ask acceleration is zero in which time . so,
2a - 6bt = 0

6bt = 2a

t = a/3b

hence, option (3) is correct .