Question

&.. The position of the moving particle to given by
x=6+3t+9t2
what is the velocity at t=2s ​

Answer 1

Given,

                          x = 6 + 3t+ 9t²

                   Time = 2 seconds

To find the velocity of the particle, we should find the derivative of x..

                 dv =  d  (6 + 3t + 9t²)

                 dx     dx

Now we need to find the derivative of the following,

         1.   d  (6)=  0

               dx

         2.   d  (3t)=  3

               dx

         3.   d (3t²)=  2 × 3t = 6t

               dx

Therefore,

                   d  (6 + 3t + 9t²) = 0 + 3+ 6t

                   dx

Now it is given that t= 2

Thus,

          6t + 3 =12 + 3 = 15

The velocity of the particle is 15

I hope my answer is correct..

Thank u and have a great day....

         

Answer 2

Answer:

The position of the moving particle to given by $x=6+3t+9t^2$

the velocity at t=2s ​ will be $39m/s$

Explanation:

• The velocity can be defined as the rate of change of position with respect to time and

• in the same way, acceleration can be defined as the rate of change of velocity or the second derivative of position with respect to time.

• Let t be the time, x be the position, v be the velocity and a be the acceleration of an object.

        Then, by definition,

        $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{d} t}$

        and

        $a=\frac{d v}{d t}$    

• In the question, it is given that,

       $x=9t^2+3 t+6$       $t= 2s$          

 

• Taking the derivative of x,

      $v=\frac{d (9t^2 +3t+6) }{d t}= 18t+3$

      put $t=2s$

       $v = (18\times 2) +3$

           $=39m/s$