In a women\'s 100-m race, accelerating uniformly, Laura takes 1.90 s and Healan

Question

In a women's 100-m race, accelerating uniformly, Laura takes 1.90 s and Healan 3.10 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? aLaura = 1 m/s2 aHealan = 2 m/s2
(b) What are their respective maximum speeds vLaura,max = 3 m/s vHealan,max = 4 m/s
(c) Which sprinter is ahead at the 5.90-s mark, and by how much? 5---Select---LauraHealan is ahead by 6 m.
(d) What is the maximum distance by which Healan is behind Laura?
7 m

At what time does that occur?
8 s (a) What is the acceleration of each sprinter? aLaura = 1 m/s2 aHealan = 2 m/s2
(b) What are their respective maximum speeds vLaura,max = 3 m/s vHealan,max = 4 m/s
(c) Which sprinter is ahead at the 5.90-s mark, and by how much? 5---Select---LauraHealan is ahead by 6 m.
(d) What is the maximum distance by which Healan is behind Laura?
7 m

At what time does that occur?
8 s 5---Select---LauraHealan is ahead by 6 m. aLaura = 1 m/s2 aHealan = 2 m/s2

Explanation / Answer

0.5*a1*3.17^2 + (a1*3.17)*(10.4-3.17) = 0.5*a2*2.01^2 + a2*2.01*(10.4-2.01)

0.5*a1*3.17^2 + (a1*3.17)*(10.4-3.17) = 100

so,

a2 =5.296 m/s2 <-----------acceleration of Laura

a1 = 3.58 m/s2 <------------acceleration of Healan

b)

max speed of Laura = 5.296*2.01 =10.64 m/s

max speed of Healan = 3.58*3.17 = 11.45 m/s

c)

after 5.9s ,

Laura is at = 0.5*5.296*2.01^2 +11.24*(5.9-2.01) = 54.4 m

Healan is at = 0.5*3.58*3.17^2+11.45*(5.9-3.17) = 49.24 m

So, Laura is ahead by = 54.4-49.24 = 5.16 m

d)

5.296*t
max distance =0.5*5.296*2.01^2 -0.5*3.58*2.01^2 = 3.47 m

at t = 2.01 s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.