a sprinter who is running a 250 race accelerates from the rest at 7.5 m/s^2 for 1.2s and maintains this speed for the remainder of the race
Answers
Answer:
time for the race is 28.37 s.
The total time is t(1) for acceleration plus t(2) for the balance of the race.
After accelerating at 7.5 m/s^2 “a” for time t = 1.2 s, the runner is running at
v = a•t = 7.5 m/s^2•1.2 s = 9 m/s.
The “average” velocity during this period of acceleration is (9 m/s - 0 m/s)/2 = 4.5 m/s. The distance d covered is
d = v(avg)•t = 4.5 m/s • 1.2 s = 5.4 m.
t(1) is 1.2 s for this 5.4 m.
The remaining distance is 250 m - 5.4 m = 244.6 m. Running at 9 m/s, the time to finish the race is
t(2) = d/v = 244.6 m / 9 m/s = 27.17 s.
The total time t(1) + t(2) is 1.2 s + 27.17 s = 28.37 s.
The average velocity for the total distance is
v(avg) = 250 m / 28.37 s = 8.812 m/s.
Answer:
Her time for the race is 28.37 s.
The total time is t(1) for acceleration plus t(2) for the balance of the race.
After accelerating at 7.5 m/s^2 “a” for time t = 1.2 s, the runner is running at
v = a•t = 7.5 m/s^2•1.2 s = 9 m/s.
The “average” velocity during this period of acceleration is (9 m/s - 0 m/s)/2 = 4.5 m/s. The distance d covered is
d = v(avg)•t = 4.5 m/s • 1.2 s = 5.4 m.
t(1) is 1.2 s for this 5.4 m.
The remaining distance is 250 m - 5.4 m = 244.6 m. Running at 9 m/s, the time to finish the race is
t(2) = d/v = 244.6 m / 9 m/s = 27.17 s.
The total time t(1) + t(2) is 1.2 s + 27.17 s = 28.37 s.
The average velocity for the total distance is
The average velocity for the total distance isv(avg) = 250 m / 28.37 s = 8.812 m/s.