Question

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

Answer 1

Answer:

v= - 27 m/s

Explanation:

Given that

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

As we know that velocity given as

$v=\dfrac{ds}{dt}$

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

As we know that acceleration given as

$a=\dfrac{dv}{dt}$

$v=\dfrac{d^2s}{dt^2}$

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