In a women\'s 100-m race, accelerating uniformly, Laura takes 1.81 s and Healan

Question

In a women's 100-m race, accelerating uniformly, Laura takes 1.81 s and Healan 3.06 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?

(b) What are their respective maximum speeds

(c) Which sprinter is ahead at the 5.90-s mark, and by how much?

(d) What is the maximum distance by which Healan is behind Laura?
m

At what time does that occur?
s

Explanation / Answer

here,

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(1.81)^2 + [(1.81)a](8.59)

Solve for a:

a = 5.81 m/s^2

Vmax = (5.81)(1.81) = 10.51 m/s

So Laura's acceleration is 5.81 m/s^2; Maximum speed = 10.51 m/s

Helen:

100 m = (1/2)a(3.06)^2 + [(3.06)a](7.34)

a = 3.68 m/s^2

Vmax = (3.68 m/s^2)(3.06 s) = 11.26 m/s

So Helen's acceleration is 3.68 m/s^2; Maximum speed is 11.26 m/s

C)

At 5.9 s:

Laura: x = (1/2)(5.81)(1.81)^2 +
(4.09)(10.51) = 52.5 m

Helen: x = (1/2)(3.68)(3.06)^2 + (2.84)(11.26) = 49.21 m

So Laura is ahead by 3.29 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:

10.51 m/s = (3.68)t

Solve for t: t = 2.86 seconds

Distance that each travels during this time:

Laura: (1/2)(5.81)(1.81)^2 + (1.05)(10.51) = 20.55 m

Helen: (1/2)(3.68)(2.86)^2 = 15.05 m
Difference = 5.5 m

So the maximum distance occurs at 2.86 seconds-
Helen is 5.5 m behind

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