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?
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
I have the answer to this question and its 35.7m. I don't understand how it is achieved through these calculations..
well, it left out the original 100 meters and we differ by a decimal point
To find the displacement of the runner, we can use the equation of motion that relates initial velocity (vₒ), final velocity (v), acceleration (a), and displacement (x):
v² = vₒ² + 2a(x - xₒ)
Let's substitute the given values into the equation:
v = 0.500 m/s (final velocity)
vₒ = 2.29 m/s (initial velocity)
a = -0.700 m/s² (acceleration)
xₒ = 100 m (initial position)
Now we can solve for x:
0.500² = 2.29² + 2(-0.700)(x - 100)
0.250 = 5.2441 - 1.4(x - 100)
0.250 - 5.2441 = -1.4x + 140
-4.9941 = -1.4x + 140
Rearranging the equation:
-1.4x = -144.9941
x = (-144.9941)/(-1.4)
x ≈ 103.567 m
Therefore, the displacement of the runner is approximately 103.567 meters.