The position of a particle moving along a straight line at any time t is given by s(t)=t^3+9t^2-27.
find the velocity of the particle at the time when the acceleration is zero.

Given:

As we know,

The velocity:

→

  ...(eqn 1)

and,

The acceleration:

→

When, a = 0

→

By putting the values in "eqn 1", we get

→

Thus the above approach is appropriate.  

Learn more about acceleration here:

link

v= - 27 m/s

Explanation:

Given that

s= t³-  9 t²-27       ( Correct from sources)

As we know that velocity given as

v=3 t ² - 18 t               ------------1

As we know that acceleration given as

v=3 t ² - 18 t

a=6 t  -18

Given that acceleration is zero (a= 0 )

0 = 6 t  - 18

t=  3 sec

Now by putting the values in the equation 1

v=3 t ² - 18 t  m/s  

v=3 x 3 ² - 18 x 3 m/s

v= 27 - 54  m/s

v= - 27 m/s

this is a good question! i'll do some research and come back when i have the answer!