a(t)=3t^2 + 2t

Question

Take the above function. If x=1, the velocity is 0 and position is -22. 

FInd v(t) at any time.

Find x(t) at any time.

 

I have no clue how to answer this. Absolutely no clue.

Answer 1

The acceleration a(t) is the derivative of the velocity function, v(t).

 

So, v'(t) = a(t) = 3t²+2t  and v(1) = 0

 

Integrate to get v(t) = t³ + t² + C₁

 

       Since v(1) = 0, 1³ + 1²+ C₁ = 0  So, C₁ = -2

 

       v(t) = t³ + t² - 2

 

The velocity function is the derivative of the position function, s(t).

 

So, s'(t) = v(t) = t³ + t² - 2 and s(1) = -22

 

Integrating, we have s(t) = ¼t⁴ + (1/3)t³ - 2t + C₂

 

   Since s(1) = -22, 1/4 + 1/3 - 2 + C₂ = -22

 

                                                 C₂ = -22 - 7/12 = -22 7/12

 

Therefore, s(t) = (1/4)t⁴ + (1/3)t³ - 2t - 22 7/12