Please explain/show work 51. In a women\'s 100-m race, accelerating uniformly, L

Question

Please explain/show work

51. In a women's 100-m race, accelerating uniformly, Laura takes 2.00 s and Healan 3.00 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 orld record of 10.4 s. (a) What is the acceleration of each sprinter? (b) What are their respective maximum speeds? (c) Which sprinter is ahead at the 6.00-s mark, and by how much? (d) What is the maximum distance by which Healan is behind Laura, and at what time does that occur?

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(2.00)^2 + [(2.00)a](8.4)
100 m = 2a + 16.8a
Solve for a: a = 5.319 m/s^2
Vmax = (5.319)(2.00) = 10.638 m/s

So Laura's acceleration is 5.319 m/s^2;

Maximum speed = 10.638 m/s


Helen:
100 m = (1/2)a(3.00)^2 + [(3.00)a](7.4)
100 m = 4.5 a + 51.8 a
a = 3.745 m/s^2
Vmax = (3.745 m/s^2)(3.00 s) = 11.236 m/s

So Helen's acceleration is 3.745 m/s^2; Maximum speed is 11.236 m/s

C)
At 6.00 s:
Laura: x = (1/2)(5.319)(2.00)^2 + (4.00)(10.638) = 53.19 m
Helen: x = (1/2)(3.745)(3.00)^2 + (3.00)(11.236) = 50.56 m

So Laura is ahead by 2.63 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:
10.638 m/s = (3.745)t
Solve for t: t = 2.84 seconds
Distance that each travels during this time:
Laura: (1/2)(5.319)(2.00)^2 + (0.84)(10.638) = 19.574 m
Helen: (1/2)(3.745)(2.84)^2 = 15.103 m
Difference = 4.471 m

So the maximum distance occurs at 2.84 seconds-
Helen is 4.471 m behind

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