A runner follows a straight line path. He reaches the 100m mark with a velocity of 2.29m/s, then slows down steadily (-0.0700m/s^2) until he reaches a velocity of (0.50m/s). What is the displacement of the runner?
To find the displacement of the runner, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
Let's plug the given values into the equation:
v^2 = u^2 + 2as
Final velocity (v) = 0.50 m/s
Initial velocity (u) = 2.29 m/s
Acceleration (a) = -0.0700 m/s^2 (negative because it is slowing down)
0.50^2 = 2.29^2 + 2 * (-0.0700) * s
0.25 = 5.2441 - 0.1400s
0.1400s = 5.2441 - 0.25
0.1400s = 4.9941
Now, solve for s (displacement):
s = 4.9941 / 0.1400
s ≈ 35.6714 meters
Therefore, the displacement of the runner is approximately 35.6714 meters.