the top sprinter in the world can reach a terminal speed of 50km/h, in 20m. Assuming the acceleration to be constant, compute the acceleration such an individual experiences, and compare it (in percentage form) to the acceleration of gravity.
V = 50,000m/3600s = 13.9 m/s.
t = 20min ? = 1200 s.
a = (V-Vo)/t =
Vo = 0.
a/g * 100% =
1.15%
ITS 20 METRES NOT MINUTES M=METRES NOT MINUTES MIN=MINUTES
I GOT AN EXAM QUESTION WRONG BECAUSE OF YOU !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IT IS METRES
To compute the acceleration experienced by the sprinter and compare it to the acceleration of gravity, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (terminal speed)
u = initial velocity (0, as the sprinter starts from rest)
a = acceleration
s = distance traveled (20m)
We know that the terminal speed of the sprinter is 50 km/h, which can be converted to meters per second (m/s) by dividing by 3.6:
v = 50 km/h = (50 * 1000) / 3600 = 13.89 m/s
Substituting the values into the equation:
13.89^2 = 0^2 + 2a * 20
193.21 = 40a
Dividing both sides by 40:
a = 4.83 m/s^2
Now, to compare this acceleration to the acceleration of gravity, which is approximately 9.8 m/s^2, we can calculate the percentage:
(acceleration of the sprinter / acceleration of gravity) * 100
(4.83 m/s^2 / 9.8 m/s^2) * 100 ≈ 49.29%
Therefore, the acceleration experienced by the sprinter is approximately 49.29% of the acceleration of gravity.