Abird experiences an acceleration of -1.80 m/s^2 for 3.81 s, and ends up 26.5 m to the left of its starting point. what is its final velocity?

To solve this problem, we can assume that the motion is linear so that we can use the equations:

Xf = Xi + v0 t + 0.5 a t^2                  ---> 1

vf = v0 + a t                                         ---> 2

where Xi = initial distance = 0, v0 = initial velocity, Xf = final distance = 26.5 m, a = acceleration = - 1.80 m/s^2, t = time = 3.81 s, vf = final velocity

Substituting the given values into equation 1 to find for v0:

26.5 = v0 (3.81) + 0.5 (-1.80 m/s^2) (3.81)^2

3.81 v0 = 39.56449

v0 = 10.38 m/s

Using equation 2 to find for vf:

vf = 10.38 + (-1.80) (3.81)

vf = 3.53 m/s