Question

position of a moving body is given by an equation X= t³+4t²-3t+5. Find the velocity and acceleration at t=2s​

Answer 1

25m/s

Explanation:

by takinh derivative of position we get velocity

by solving this we obtain 25m/s

Answer 2

Answer:

The position of a moving body is given by the equation X= t³+4t²-3t+5.

The velocity and acceleration at t=2s will be $25m/s$ and$20m/s^2$ respectively.

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,

       $x=t^{3}+4 t^{2}-3 t+5$          $t=2s$

 

• Taking the derivative of x,

      $v=\frac{d (t^3+4t^2-3t+5) }{d t}= 3t^2+8t-3$

      put $t= 2s$  

       $v=( 3\times 2^2)+(8\times2)-3$

           $=12+16-3= 25m/s$

• Taking the derivative of v

        $a=\frac{d (3t^2+8t-3) }{d t}= 6t+8$

       put $t= 2s$ to obtain acceleration at t=2s

       $a = (6\times 2)+8 = 20m/s^2$