In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 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) which sprinter is ahead at the 6.15-s mark, and by how much?
b)what is the maximum distance by which Healan is behind Laura?
c) At what time does that occur?

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m. In all the races the rest starts, so the initial speed is zero (Vo = 0)

Vf1= Vo + a1 t1

Vf1 = x/t

XT = X1 + X2

X1 = Vo t1 + ½ a1 t1²

X1 = ½ a1 t1²

X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

a1= XT / [ ½ t1² + t1 (t-t1)]

a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

a1= 5.79m/s2

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2

Vf2 = x/t

XT = X3 + X4

X3 = Vo t2 + ½ a2 t2²

X3 = ½ a2 t2²

X4 = Vf2 (t-t2)

XT = ½ a2 t2² + Vf2 (t- t2)

XT = ½ a2 t2² + a2 t2 (t-t2)

The aceleration of Healan (a2)

a2 = XT / [½ t2² + t2 (t-t2)]

a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]

a2 = 3.67 m / s2

We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

Vf1 = 0 + 5.79 1.82

Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

Vf2 = 0 + 3.67 3.07

Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)

XL= 55.23 m

Healan

XH= ½ a2 t2² + Vf2 (t3-t2)

XH= ½ 3.67 3.07² + 11.27 (6.15-3.07)

XH= 52.01 m

(XL -XH)= 55.23- 52.01

(XH -XL)= 3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

XL= ½ a1 t12

XL= ½ 5.79 1.822

XL= 9.59 m

XH = ½ a2 t12

XH= ½ 3.67 1.822

XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units