Sunday, July 13, 2014

A runner hopes to complete the 10,000 m run in less than 30.0 min. After exactly 27.0 min, there are still 1100 m to go. The runner must then...

There are two parts to the problem: the runner will accelerate for a time duration and then will run at a constant rate, in order to get to the target in 30 minutes. 


Assuming that the runner ran at a uniform speed for first 27 minutes, his speed is calculated as:


Speed for 27 minutes = distance / time = (10000 - 1100) m / (27 x 60) s = 5.494 m/s.


Thus, u = 5.494 m/s


acceleration, a = 0.2 m/s^2


If he accelerates for t seconds, then v = u + at = 5.494 + 0.2 x t m/s


Now, the remaining 1100 m is ran in two stretches: accelerating stretch and constant speed stretch, both combined for the remaining 3 minutes (= 180 seconds)


The distance covered in accelerating stretch is equal to ut + 1/2 at^2 m


distance ran during constant speed stretch is equal to v x (180 - t) m


Thus, ut + 1/2 at^2 + v x (180 - t) = 1100 


substituting, v = u + at and also the values of u and a and solving the equation, we get:


t = 3.1125 s or 356.8875 s


Since, an answer greater than 180 s is not possible, the time is approximately 3.1 s.


Thus, the runner needs to accelerate for about 3.1 s and then run at a constant speed to get to the target in desired time frame.


Hope this helps. 

No comments:

Post a Comment

What are hearing tests?

Indications and Procedures Hearing tests are done to establish the presence, type, and sever...