Starting from the constant-acceleration kinematic equations, write a formula that gives xf in terms of t, xi, vf, and a.

Huh ?

in terms of vf?????

yes in terms of vf. we are asked to combine different kinematic equations to drive xf. I believe the equation v=vo+at^2 can sub for vo. does this make sense?

v = Vi + a t

vf = Vi + a T so Vi = vf-aT

x = xi + Vi t + (1/2) a t^2
xf = xi +(vf-aT)t + (1/2) a t^2
if t = T final time
xf = xi + vf T - a T^2 + (1/2) a T^2

xf = xi + vf t - (1/2) a t^2

ok, it just changes the sign of the at^2 term
(It still uses the AVERAGE speed during the interval !!)

To derive the formula for xf (final position) in terms of t (time), xi (initial position), vf (final velocity), and a (acceleration) using the constant-acceleration kinematic equations, follow these steps:

Step 1: Start with the formula:
xf = xi + vit + (1/2)at^2

Step 2: Rearrange the equation to isolate xf on one side:
xf - xi = vit + (1/2)at^2

Step 3: We can now apply another kinematic equation that directly relates velocity and time. The equation is:
vf = vi + at

Step 4: Rearrange the equation from Step 3 to isolate vi:
vi = vf - at

Step 5: Substitute the value of vi from Step 4 into the equation in Step 2:
xf - xi = (vf - at)t + (1/2)at^2

Step 6: Expand the equation further:
xf - xi = vft - at^2 + (1/2)at^2

Step 7: Simplify the equation:
xf - xi = vft - (1/2)at^2

Step 8: Finally, isolate xf by bringing xi to the other side:
xf = xi + vft - (1/2)at^2

Therefore, the formula to calculate xf in terms of t, xi, vf, and a is:
xf = xi + vft - (1/2)at^2