In a women\'s 100-m race, accelerating uniformly, Laura takes 1.86 s and Healan

Question

In a women's 100-m race, accelerating uniformly, Laura takes 1.86 s and Healan 2.99 s to attain their maximum speeds, which they each maintain for the rest of the race. They cross the finish line simultaneously, both setting a world record of 10.4 s. (a) What is the acceleration of each sprinter? a_Laura = m/s^2 a_Healan = m/s^2 (b) What are their respective maximum speeds V_Laura, max = m/s V_Healan, max = m/s (c) Which sprinter is ahead at the 6.40-s mark, and by how much? is ahead by m. (d) What is the maximum distance by which Healan is behind Laura? m At what time does that occur? S

Explanation / Answer

a] For Laura:

time taken after attaining the maximum velocity to cover the rest of the distance = 10.4 - 1.86 = 8.54 s

total distance covered = S = 100 m

so, S = ut1 + (1/2)at12 + vt2

=> 100 = 0 + (1/2)a(1.86)2 + a(1.86)(8.54)

=> a = 5.677 m/s2

Repeat the same for Healan. Acceleration for her comes out to be: a = 3.756 m/s2.

b]

so, the maximum speed for Laura will be: v = u + at1 = 0 + 5.677(1.86) = 10.56 m/s.

similarly, for Healan, maximum speed = v = 0 + 2.99(3.756) = 11.23 m/s.

c] at t = 6.4s

distance covered by Laura = S = (1/2)(5.677)(1.86)2 + 10.56(6.4 - 1.86) = 57.76 m

and for Healan: S = (1/2)(3.756)(2.99)2 + 11.23(6.4 - 2.99) = 55.08 m

therefore, Laura is ahead of Healan by d = 57.76 - 55.08 = 2.676 meters.

d]

Laura's distance: S = 9.82 + 10.56(t - 1.86)

Healan's distance: S' = 16.78 + 11.23(t - 2.99)

so, S - S' = 9.82 + 10.56(t - 1.86) - 16.78 - 11.23(t - 2.99)

=> S - S' = 6.966 - 0.67t

this distance between them will be maximum when t = 0s, and for this, maximum distance of separation = 6.966 m.

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