A car starts from rest and accelerates uniformly to a velocity of 8ft/s after traveling 250 ft. Find its acceleretaion.

Help me with this pleeaase.
Thank you.

In this situraion,

Vf = sqrt(2aX) where
Vf = the final velocity, 8 ft/s
X = the distance traveled, 250 ft
Solve for the acceleration, a, in ft/s^2.

i don't get it

What don't you get? The derivation of the first formula or how to solve it for a?

Here is how to derive the first formula.

Vf = a t
X = (1/2) a t^2

Vf^2 = a^2 t^2 = 2 a X
Vf = sqrt(2 a X)

To find the acceleration of the car, we can use the equation of motion that relates distance, initial velocity, final velocity, and acceleration.

The equation is:
v^2 = u^2 + 2as

Where:
v = final velocity = 8 ft/s
u = initial velocity = 0 ft/s (since the car starts from rest)
s = distance traveled = 250 ft
a = acceleration (what we need to find)

Plugging in the given values, we have:
8^2 = 0^2 + 2a(250)

64 = 2a(250)

Dividing both sides of the equation by 2(250), we get:
64 / 500 = a

Simplifying further, we have:
0.128 = a

Therefore, the acceleration of the car is 0.128 ft/s^2.

To solve problems like this, you can use the equation of motion and substitute the given values to find the unknown variable.