running at 10/3 mt/sec lost by 3 secs to the winner. if the runners up to 15 secs to cover the distance, what was the speed of the winner?
Answers
You are using an average rate of speed to represent a smaller leg of the trip. In the formula:
Which is a basic kinematics formula for acceleration and deceleration, S represents your position, V avg is your average velocity, t is time, V0 is initial velocity, V is final velocity, and a is acceleration.
So, if you start from a standstill, your V0 = 0 m/s. If we take the 2nd and third portions of the equation as equal, then Vavg= (V0+V)/2, and further, if we take your 100 m in 10 seconds and calculate this as a velocity of 100m/10s = 10m/s, we see that since we started from a complete stop, that this speed must represent our average velocity.(Our final velocity must be double this 10m/s for the math to work, so at the fastest, we are traveling 20m/s!!)
Now, look at the last segment of the formula. In a race, a sprinter does not actually run only 100 meters when they run a 100 m dash. They run 100 meters full speed, and then they don’t even think about putting on the brakes until they are slightly past the finish line, so during the part that was timed, there is only an acceleration, and no deceleration. The fastest, then, a track athlete can run is to give himself a high acceleration, and maintain it over the entire course of the race, so he reaches max velocity right at the finish line.( I use “he” for the athlete’s gender here for simplicity, there are many fine female sprinters out there that are equally talented track runners, and I seriously doubt I could beat either of my sister’s in a 100 m sprint, as they were div II sprinters in college, even if I ran track in high school) If this is the case, there is never a steady state, so you can use the fourth section of the formula for the entire length of the race, rather than having to add a steady speed segment to the formula.(e.g. the runner runs at a steady state after 3 seconds of acceleration, you would use the formula for the first 3 seconds, then use Vs*t for the rest of the race, where Vs is the steady state velocity, as a piecewise function.)
The fourth section of the formula simplifies to (a*t^2)/2 =(V*t)/2=S, so, you can calculate that for a constant acceleration, 100 m, run in 10 s, with a final velocity of 20 mm/s, the acceleration would be 2mm/s^2. You can then build a chart of the average speed at each 1 second interval:
In this simple case, you are only running under your 10 m/s speed for the first 5 seconds, at which point you continue to run faster than that!!!
Hopefully this proves that you can run this fast, albeit for a short run. Now obviously, as a runner moves, they are not reaching a constant acceleration, and there are various gradual changes in velocity and acceleration during a run.(i.e. a person is not on wheels, so the rate comes in small surges with each footfall, and a runner might stride over a meter with each step, so it is more a function of getting your feet to hit the ground rapidly, while also making the longest extension possible in that small amount of time, but the simple math at least illustrates that it is possible)