Acar slows down uniformly from a speed of 29.0 m/s to rest in 8.20 s how far did it travel

For this problem, we use the derived equations for rectilinear motion at constant acceleration.

a = (v - v₀)/t
x = v₀t + 0.5t²
where
a is acceleration
v₀ and v are the initial and final velocities, respectively
x is the distance
t is the time

The solution is as follows:

a = (29 m/s - 0 m/s)/8.20 s = 3.54 m/s²
Using this to the second equation:
x = (0 m/s)(8.20 s) + 0.5(3.54 m/s²)(8.20 s)²
x = 119 m

answer:

y = 3 - 1

step-by-step explanation:

-1 is the intercept on the y axis and 3 is up 3 over 1