Question

If a position of a object is 2t^3 +3t^2+ t find velocity and acceleration

Answer 1

Answer:

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Answer 2

Answer:

If a position of an object is 2t^3 +3t^2+ t,

• The velocity will be $6t^2+6t+1$

• The acceleration will be  $12t+6$

Step-by-step explanation:

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

• In the question, it is given that,

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

• 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

        $a=\frac{d v}{d t}$

• Taking the derivative of x,

       $v=\frac{d x}{d t}= 6t^2+6t+1$

       Then again taking the derivative of v,

        $a=\frac{d v}{d t}=12t+6$