## anonymous one year ago Help!

1. anonymous

WOULDN'T YOU JUST PLUG THEM IN? $3(3)^{2}-4(3)+1$

2. anonymous

No because thats not an option

3. Michele_Laino

you have to compute the first derivative of the function v(t) at t=3, namely: $\Large a\left( t \right) = \frac{d}{{dt}}\left( {3{t^2} - 4t + 1} \right) = ...?$

4. anonymous

6t-4

5. Michele_Laino

ok! now replace t with 3, what do you get?

6. anonymous

14

7. Michele_Laino

8. anonymous

Thank you so much!

9. anonymous

michele is right cause accelration is the rate of change of velocity per time!

10. Michele_Laino

thanks!! :)

11. Michele_Laino

thanks!! @Hassan3130

12. anonymous

Thank you @Hassan3130 !