AP Calculus AB- Constant of Proportion for Acceleration?

[AndrewHelmerich]

I know that the derivative of displacement is velocity, the acceleration; vice versa. I asked my teacher about this and he said that we cannot use any of the physics equations (vf=vi+at, etc). Can some one help? It would be much appreciated.

 

[skeeter]

[MATH]a(t) = kt^2[/MATH]
$$v(t) = \dfrac{kt^3}{3} + C$$
$$v(0) = 324 \implies v(t) = \dfrac{kt^3}{3} + 324$$
[MATH]v(t_f) = 0 \implies \dfrac{kt_f^3}{3} = -324[/MATH] (1)

[MATH]x(t) = \dfrac{kt^4}{12} + 324t + C[/MATH]
[MATH]x(0) = 0 \implies C=0[/MATH]
$$x(t_f) = \dfrac{kt_f^4}{12} + 324t_f = 1039.5$$ (2)

two equations & two unknowns

 

[AndrewHelmerich]

[MATH]a(t) = kt^2[/MATH]
$$v(t) = \dfrac{kt^3}{3} + C$$
$$v(0) = 324 \implies v(t) = \dfrac{kt^3}{3} + 324$$
[MATH]v(t_f) = 0 \implies \dfrac{kt_f^3}{3} = -324[/MATH] (1)

[MATH]x(t) = \dfrac{kt^4}{12} + 324t + C[/MATH]
[MATH]x(0) = 0 \implies C=0[/MATH]
$$x(t_f) = \dfrac{kt_f^4}{12} + 324t_f = 1039.5$$ (2)

two equations & two unknowns

Thank you, I appreciate it.