Aparticle moves with position function1
s = t
4 − 4t
3 − 20t
2 + 20t t ≥ 0
(a) at what times does the particle have a velocity of 20 m/s? velocity v(t) is the derivative of the position
function s(t) (b) at what time is the acceleration 0? what is the significance of this value of t?

(a): The velocity of the particle will be 20 m/s at 0 second and 5 seconds.

(b): The acceleration of the particle will be zero at 3.08 seconds.

Significance of this time: The velocity of the particle is minimum at this point.

Explanation:

The velocity of the partcle at an instant of time t is equal to the derivative of the position of the particle at that time.

The accelartion of the particle at an instant of time t is equal to the derivative of the velocity of the particle at that time.

Given that the position of the particle is

The velocity of the particle is given by

The velocity of the particle will 20 m/s when,

The accelartion of the particle is given by

The acceleration is 0 when,

Since, the acceleration of the particle at this time is 0, i.e., the acceleration before this time would be negative and it is positive after this time.

It means that the velocity of the particle before this time is decreasing and that after this time is increasing. Thus, the velocity of the particle is minimum at this point.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach