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.