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.
Find the velocity of the particle at the time when the acceleration is zero.

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

   

Answer 2

The velocity of the particle will be "-27 m/s".

Given:

• $s(t)=t^3+9t^2-27$

As we know,

The velocity:

→ $v = \frac{ds}{dt}$

  $v = 3t^2 -18 t$ ...(eqn 1)

and,

The acceleration:

→ $a = \frac{dv}{dt}$

  $a = 6t -18$

When, a = 0

→ $0 = 6t -18$

 $6t = 18$

   $t = \frac{18}{6}$

      $= 3 \ sec$

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

→ $v = 3t^2-18t$

     $= 3\times 3^2-18\times 3$

     $= 27-54$

     $= -27 \ m/s$

Thus the above approach is appropriate.  

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