Answer:
a) The max speed is 8 m/s
b1) Trevor will travel 24 m in 3 s after reaching max speed.
b2) After 9 s of starting the race, Trevor will run 64 m
Step-by-step explanation:
For a straight movement with constant speed (as Trevor running after 6 s) the variation of the position over time (dx / dt) is constant (v):
dx / dt = v
separating variables:
dx = v dt
Integrating both sides of the equation, the left side between the position at time "t" (x) and the initial position (x0), and the right side between time "t" and the initial time (t = 0):
x -x0 = v t
x = vt + x0
a ) We know that Trevor runs 72 m in 10 s after starting the race, 6 s at variable speed and 4 s at constant max speed. We can use the equation of position to obtain this max speed:
x = vt + x0
Since Trevor starts running at constant speed after 40 m, the initial position will be 40 m. The time will be 4 s because that is the time that Trevor runs at constant speed. Then:
72 m = v * 4 s + 40 m
v = 8 m/s
b1) Knowing the max speed, we can calculate how far Trevor will run during 3s after reaching the max speed:
x = vt = 8 m/s * 3s = 24 m
b2) During the first 6 s Trevor runs 40 m and 3 s after reaching 40 m Trevor runs 24 m more. The Trevor´s distance from the starting line after 9 s will be
40 m + 24 m = 64 m