Respuesta :
Answer:
See explanation below.
Explanation:
For this case we have the following function:
[tex] x(t)= ct^4 +dt^2 +f[/tex]
Where [tex] c= 6 m/s^4 , d = 8m/s^2 , f=-6m [/tex]
If we replace those values we got:
[tex] x(t) = 6t^4 + 8t^2 -6[/tex]
If we want to find the position after t = 2.358 seconds for example we ust need to replace in the position function t = 2.358 and we got:
[tex] x(t=2.358) = 6(2.358)^4 + 8(2.358)^2 -6 \approx 224 m[/tex]
If we want to find the velocity we need to take the derivate of the position function and we got:
[tex] \frac{dx}{dt}=v(t) = 4ct^3 + 2dt[/tex]
[tex] v(t)= 24 t^3 + 16 t[/tex]
If we want to find the instantaneous velocity we just need to replace on v(t) a value for t
And the accelaration would be given by the second derivate of the position:
[tex] \frac{dv}{dt}= 72 t^2[/tex]
If we want to find the instantaneous acceleration we just need to replace on v(t) a value for t
