How do the speeds v0, v1, and v2 (at times t0, t1, and t2) compare?

a: v0 = v1 = v2 > 0
b: v0 = v2 > v1 = 0
c: v0 = v2 > v1 > 0
d: v0 > v1 > v2 > 0
e: v0 > v2 > v1 = 0

option c: v₀ = v₂ > v₁ > 0

Explanation:

1) Velocity is the rate of change of the position with time. Therefore, in a graph of position vs time, the velocity is the slope of the path that describes the motion.

2) In this case you are requested to compare the speeds, i.e. just the magnitudes of the velocities. You can perform that by comparing the slopes in the three positions (x₀, y₀), (x₁, y₁), and (x₂, y₂).

3) Regarding v₁, since it corresponds to the peak of the curve (x₁.y₁), the vertical component of the velocity is zero (the horizontal component is constant, since there is not any horizontal force acting), so at this point the speed reaches its minimum value, then v₁ > 0, and v₁ is less than any other speed.

4) Regarding v₀ and v₂, at (x₀, y₀) and (x₂, y₂), note that those are points at the same distance from the axis of symmetry of the curve (assuming it is perfect parabola, which is what the graph intends to show).

That means that the velocities at those points are equal in magnitude but opposite in signs. Since, you are asked about the speed, which is magnitude, you conclude v₀ = v₂.

In conclusion v₀ = v₂ > v₁ > 0, which is the option c.

i think it's 35 meters