Question

the position of a particle in x axis is given by x=t^3-t^2+2t find the velocity when time is 2secs

Answer 1

Hello,

We know that,

Velocity = d/dt( displacement)

So given displacement, x=t^3-t^2+2t

d/dt(x) =3t^2-2t^1+2

Velocity at t=2 secs is given by,

(d/dt(x)) at t=2 secs

ie,d/dt(x) = 3*2^2-2*2^1+2

= 10 m/sec

Hence,

Velocity =10 m/sec.

Answer 2

Answer:

The position of a particle in the x-axis is given by $x= t^3+t^2+2t$  The velocity when time is 2secs  will  be $10m/s$

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,

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

• Taking the derivative of x,

      $v=\frac{d( t^3 -t^2+2t)}{d t}= 3t^2-2t+2$

      At time $t= 2s$  

       $v= (3\times 2^2)-(2\times 2)+2= 10 m/s$