Question

The position of an object moving along x axis is given by 3t^3+2t^2+t ,find the 'a' (acceleration) at t=2sec​

Answer 1

Answer:

The position of an object moving along the x-axis is given by $x= 3t^3+2t^2+t$ the 'a' (acceleration) at t=2sec​ is, $40m/s^2$

Explanation:

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

        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,

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

• Taking the derivative of x,

      $v=\frac{d( 3t^3+2t^2+t)}{d t}= 9t^2+4t+1$

       Again taking derivative of v,

       $a=\frac{d v}{d t}= 18t+4$

       At time $t= 2s$  

       $a= (18\times2)+4=40 m/s^2$