In a women\'s 100-m race, accelerating uniformly, Laura takes 2.10 s and Healan

Question

In a women's 100-m race, accelerating uniformly, Laura takes 2.10 s and Healan 3.21 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 5.50-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

aLaura =
Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2 aHealan =
Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2 In a women's 100-m race, accelerating uniformly, Laura takes 2.10 s and Healan 3.21 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 wor is ahead by m. (d) What is the maximum distance by which Healan is behind Laura? M a Laura At what time does that occur? s ld record of 10.4 s. (a) What is the acceleration of each sprinter? Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2 (b) What are their respective maximum speeds vLaura,max = m/s vHealan,max = m/s (c) Which sprinter is ah Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2 aHealan = ead at the 5.50-s mark, and by how much?

Explanation / Answer

Total distance s = 100 m

Time laura takes to attain max speed vmax,l : tl = 2.10 s

let the distance covered till she attains max speed = s1l

Once a sprinter attains max speed she maintains it till the rest of race.

So the race has 2 parts:

part 1: runs distance s1 in time t' with acceleration a, initial speed = 0, final speed = vmax

part 2: runs distance s2 in time t'' with acceleration 0, initial speed = v_max

So,

s1 + s2 = s

t' + t'' = t

ACCELERATION

1. For Laura:

t' = 2.1 s

s1 = 0 * t' + 1/2 a t''2 = 1/2 a*2.12 = 2.205a

vmax = 0+a*t' = a * 2.1 = 2.1a

s2 = u*t'' = vmax (10.4-2.1) = 8.3vmax = 8.3 * 2.1a = 17.43 a

Now we know s1 + s2 = s = 100m

so, we add the two:

100 = 2.205a + 17.43 a = 19.635a

so, a =100/19.635 = 5.093 m/s2

Acceleration of Laura = 5.093 m/s2

Maximum speed of Laura: 2.1a = 10.695 m/s

2. Now we find acceleration of Helean in same manner:

t' = 3.21 s

s1 = 0 * t' + 1/2 a t''2 = 1/2 a*3.212 = 5.15a

vmax = 0+a*t' = a * 3.21 = 3.21a

s2 = u*t'' = vmax (10.4-3.21) = 7.19vmax = 7.19 * 3.21a = 23.08 a

Now we know s1 + s2 = s = 100m

so, we add the two:

100 = 5.15a + 23.08 a = 28.23a

so, a =100/28.23 = 3.54 m/s2

Acceleration of Helean = 3.54 m/s2

Maximum speed of Helean: 3.21a = 11.695 m/s

QUES: Which sprinter is ahead at the 5.50-s mark, and by how much?

t = 5.5 s

Now both sprinters have stopped accelerating at this point. (as Laura accelerates till 2.1 s and Helean till 3.21 s)

We find the distance covered by both in both parts of the race:

Laura:

Till time t' = 2.1 s distance covered:

s1 = 2.205a = 2.205 * 5.093 = 11.23 m

remaining time: 5.5 - 2.1 = 3.4 s

In part 2 of race distance covered in 3.4 s:

s2 = vmax * 3.4 = 10.695 * 3.4 =36.363 m

Total distance covered by Laura in 5.5 s = 11.23 + 36.363 = 47.593 m

Similarly we find distance covered by Helean:

Till time t' = 3.21 s distance covered:

s1 = 5.15a = 5.15 * 3.54 = 18.231 m

remaining time: 5.5 - 3.21 = 2.29 s

In part 2 of race distance covered in 2.29 s:

s2 = vmax * 2.29 = 11.695 * 2.29 =26.782 m

Total distance covered by Helean in 5.5 s = 18.231 + 26.782 = 45.013 m

Thus Laura is ahead of Helean at time 5.5 s of race.

QUES: What is the max distance by which Helean is behind Laura and at what time does it occur?

From the start of the race the acceleration of Laura is more, she accelerates till t = 2.1 seconds and stops accelerating thereafter. Helean is constantly lagging behind laura till this time = 2.1 s, after that time the gap between them starts bridging as laura has stop accelerating whereas helean is still accelerating.

So maximum gap between then occurs at t=2.1 s

And the distance between them:

Laura's distance at t=2.1 : 11.23 m (see above)

Heleans's distance:

s1 = 1/2 a 2.12

Substitute Helean's acceleration:

s1 = 1/2 * 3.54 * 2.12 = 7.806 m

Thus distance between them: 11.23 - 7.806 = 3.424 m This is the max distance by which Helean is behind Laura and it accurs at t=2.1 s

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