Question

the displacement of particle starting from rest is given by x=3t^2-t^2 calculate the time at which acceleration of particle become zero

Answer 1

v= dx/dt

v= d(3t^2-t^2)/dt

v = 6t- 2t

then a= dv/dt

a=d(6t- 2t)/dt

a= 4

it is independent of time i.e. it is constant

Answer 2

Explanation:

differentiate above equation we get

V = 6t - 2t

Again differentiating

A = 6-2

A=4

Acceleration is constant and does not change

never becomes 0