Question

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?

Answer 1

Laura is ahead and for a distance of 3.22 m

Step-by-step explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

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