3. Chapter 2. TSU 50m Challenge. TSU has a rich tradition in Track and Field. In

Question

3. Chapter 2. TSU 50m Challenge. TSU has a rich tradition in Track and Field. In 1960, TSU graduaté In Wilma Rudolph won the Olympic gold medal in the women's 100 m dash with a time of 11.18 seconds. 1968, TSU graduate Madeline Manning earned Olympic gold in the 800 m run, completing it in a time o 120.9 seconds. (a) Assuming she started from rest and ran with constant acceleration over the entire 100 m, what was Wilma Rudolph's acceleration during the race? (b) Assuming she ran the entire 800 m at a constant speed (over such a long distance, the accleration portion is very small), with what constant speed did Madeline Manning run during her race? (c) Suppose a special event had been organized between the two, as follows: Madeline Manning is allowed to approach the starting line already at her top speed (as found in part b). The instant she crosses, Wilma Rudolph starts accelerating from rest at the starting line, accelerating at the value found in part a. The finish line is 50 m away. How long does it take for each runner to finish the race? Who wins? (d) As they cross the finish line, what is the final speed of each runner? Who is running faster at the end? (e) For what race distance(s) would the race end in a tie?

Explanation / Answer

(a) Distance s = 100 m

time of finish, t = 11.18 s

use the expression –

s = u*t + (1/2)*a*t^2

=> 100 = 0 + 0.5*a*11.18^2

=> a = 100 / (0.5*11.18^2) = 1.6 m/s^2

(b) distance s = 800 m

time t = 120.9 s

speed = s / t = 800 / 120.9 = 6.62 m/s

(c) Time taken by Madeline Manning, t1 = 50 / 6.62 = 7.55 s

time taken by Wilma Rudolph –

50 = 0 + (1/2)*1.6*t^2

=> t = sqrt(100 / 1.6) = 7.9 s

So, Madeline Manning will win the race.

(d) suppose s is the requisite distance.

so –

s / 6.62 = sqrt[2s/1.6]

=> s^2 / 6.62^2 = 2s / 1.6

=> s / 6.62^2 = 2 / 1.6

=> s = (2 x 6.62^2) / 1.6 = 54.8 meter.

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