In a women\'s 100-m race, accelerating uniformly, Laura takes 2.01 s and Healan

Question

In a women's 100-m race, accelerating uniformly, Laura takes 2.01 s and Healan 3.23 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 5.65-s mark, and by how much?

---Select--- Laura Healan 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 and b)

for Laura:
Vmax = Vi+a*t1= a*t1 = a*2.01
Total distance = 0.5*a*t1^2 + Vmax* (t2-t1)
=0.5*a*2.01^2 + Vmax * (10.4-2.01)
= 0.5*a*2.01^2 + a*2.01 * (10.4-2.01)
= 2.02 a + 16.86a
100 = 18.88 a
a = 5.3 m/s^2
Vmax = a*2.01 = 5.3*2.01 = 10.6 m/s

for healen:
Vmax = Vi+a*t1= a*t1 = a*3.23
Total distance = 0.5*a*t1^2 + Vmax* (t2-t1)
=0.5*a*3.23^2 + Vmax * (10.4-3.23)
= 0.5*a*3.23^2 + a*3.23 * (10.4-3.23)
= 5.22 a + 23.16a
100 = 28.38 a
a = 3.52 m/s^2
Vmax = a*3.23 = 3.52*3.23 = 11.36 m/s

C)
At 5.6 s
d of Laura = 0.5*a*t1^2 + Vmax* (t2-t1)
=0.5*5.3*2.01^2 + 10.6 * (5.65-2.01)
= 49.3 m
d of Haelen = 0.5*a*t1^2 + Vmax* (t2-t1)
=0.5*3.52*3.23^2 + 11.36* (5.65-3.23)
= 45 .8 m
Laura is ahead by (49.3-45.8) = 3.5 m

d)
At any time t:
d of Laura = 0.5*a*t1^2 + Vmax* (t2-t1)
=0.5*3.52*3.23^2 + 10.6* (t-2.01)
=10.7 + 10.6t - 21.3
= -10.6 + 10.6 t

d of Haelen = 0.5*a*t1^2 + Vmax* (t2-t1)
=0.5*3.52*3.23^2 + 11.36* (t-3.23)
= 18.36 + 11.36 t - 36.7
= -18.34 +11.36 t
laura is ahead by = -10.6 + 10.6 t + 18.34 - 11.36 t = 7.74 - 0.76 t

This part is becoming very lengthy. Please do it by yourself or post it as different question

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