Answer:
The max speed is 8m/s, 3 seconds at this speed gives 24mts, at 9 seconds has traveled 64mts and at 7,5 52mts
Explanation:
The max speed can be calculated since, at 6 seconds is at 40 mts from the starting point and at constant speed is at 72 mts 10 seconds after the start of the race, then the constant max speed is (72-40)mts /(10-6)seconds=8 m/s.
the distanced travel at this speed is obtained by multipliying by 3 seconds. that is 8m/s*3s=24 mts. the distance at 9 seconds is 40mt of initial 6 plus 24 additional, that is 64mts and at 7,5 is the 40mts initial plus the 13 additional from 8m/s*1,5s = 12mt, that is 52mts.
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
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
