Question

A runner follows a straight line path. He reaches the 100m mark (xₒ = 100m) with a velocity of 2.29m/s, then slows down steadily (a = -0.700m/s^2) until he reaches a velocity of 0.500m/s. What is the displacement of the runner?

Answer 1

well 100 m plus his average speed * time during the slowdown.
to get that
v = Vi + a t
0.500 = 2.29 - 0.700 t
t = 2.56 seconds
average speed during slowdown = (2.29 + .5)/2 = 1.4 m/s
1.4 m/s * 2.56 seconds =3.58 meters more

Answer 2

I have the answer to this question and its 35.7m. I don't understand how it is achieved through these calculations..

Answer 3

well, it left out the original 100 meters and we differ by a decimal point