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

ID: 1655156 • Letter: I

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) Distance covered = (1/2)at^2 + (at)(10.4 - t)
Where t = time that it takes to accelerate to maximum speed; (at) = maximum speed
So (10.4 - t) = time that they each run at their maximum speed

Laura:
100 m = (1/2)a(1.86)^2 + [(1.86)a](10.4-1.86)
100 m = 1.73a + 15.88a
Solve for a: a = 5.678 m/s^2
Vmax = (5.678)(1.86) = 10.56 m/s

So Laura's acceleration is 5.678 m/s^2; Maximum speed = 10.56 m/s


Helen:
100 m = (1/2)a(2.99)^2 + [(2.99)a](10.4-2.99)
100 m = 4.47 a + 22.15 a
a = 3.755 m/s^2
Vmax = (3.755 m/s^2)(2.99 s) = 11.23 m/s

So Helen's acceleration is 3.755 m/s^2; Maximum speed is 11.23 m/s
b) Laura : Vmax = (5.678)(1.86) = 10.56 m/s

Helen : Vmax = (3.755 m/s^2)(2.99 s) = 11.23 m/s
c)
At 6.40 s:
Laura: x = (1/2)(5.678)(1.86)^2 + (4.54)(10.56) = 57.76 m
Helen: x = (1/2)(3.755)(2.99)^2 + (3.41)(11.23) = 55.079 m

So Laura is ahead by 2.68 m

d)
Maximum distance between runners occurs when each has the same velocity
Setting the two equal to each other:

Laura has already reached her maximum speed while Helen is still accelerating so:
10.56 m/s = (3.755)t
Solve for t: t = 2.81 seconds
Distance that each travels during this time:
Laura: (1/2)(5.678)(1.86)^2 + (0.95)(10.56) = 20 m
Helen: (1/2)(3.755)(2.81)^2 = 14.825 m
Difference = 5.175 m

So the maximum distance occurs at 2.81 seconds-
Helen is 5.175 m behind

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