Question

A particle position in meters is given by the function x(t) = ct^4 + dt^2 + f. where c = 6m/s^4, d=8m/s^2, and f=-6m and t is in seconds.

Answer 1

Answer:

See explanation below.

Explanation:

For this case we have the following function:

$x(t)= ct^4 +dt^2 +f$

Where $c= 6 m/s^4 , d = 8m/s^2 , f=-6m$

If we replace those values we got:

$x(t) = 6t^4 + 8t^2 -6$

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:

$x(t=2.358) = 6(2.358)^4 + 8(2.358)^2 -6 \approx 224 m$

If we want to find the velocity we need to take the derivate of the position function and we got:

$\frac{dx}{dt}=v(t) = 4ct^3 + 2dt$

$v(t)= 24 t^3 + 16 t$

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:

$\frac{dv}{dt}= 72 t^2$

If we want to find the instantaneous acceleration we just need to replace on v(t) a value for t