In a women's 100-m race, accelerating uniformly, laura takes 2.14 s and healan 3.15 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 = m/s2 ahealan = m/s2
(b) what are their respective maximum speeds vlaura, max = m/s vhealan, max = m/s
(c) which sprinter is ahead at the 5.85-s mark, and by how much? is ahead by m.
(d) what is the maximum distance by which healan is behind laura? m at what time does that occur? s

a) aLaura = 5.01 m/s²

aHealan = 3.60 m/s²

b) v Laura,max = 10.7 m/s

v Helana,max = 11.3 m/s

c) Laura is ahead by 2.8 m.

d) Healan is behind Laura for 98.7 m  (10.3 s).

Explanation:

The equations for position and velocity of the runners at time "t" while running with constant acceleration are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the runner at time "t".

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity at time "t".

Since the runners start from rest at the origin of the frame of reference ( the start line), x0 and v0 = 0

a)When the runners run at constant velocity their position will be:

x = x0 + v · t

Both runners reach the 100 m at 10.4 s.

For Laura, the time traveled at constant speed is (10.4 s - 2.14 s) 8.26 s

For Healan, (10.4 s - 3.15 s) 7.25 s

The total traveled distance will be the sum of the distance traveled with constant acceleration plus the distance traveled at constant velocity.

Be:

x₁, v₁, a₁ = position, velocity and acceleration of Laura respectively.

x₂, v₂, a₂ = position, velocity and acceleration of Healan respectively.

At t = 10.4 s:

x₁ = x₂ = 100 m

x0₁ + v₁ · 8.26 s = 100 m

x0₂ + v₂ · 7.25 s = 100 m

The initial position of Laura will be the position reached after 2.14 s:

x0₁ = 1/2 · a₁ · (2.14 s)²        

and her velocity will be the velocity reached during that 2.14 s:

v₁ = a₁ · 2.14 s

Replacing v₁ and x0₁ in the equation of position at t = 10.4 s:

1/2 · a₁ · (2.14 s)²  + a₁ · 2.14 s · 8.26 s = 100 m

a₁ · (1/2· (2.14 s)² + 2.14 s · 8.26 s) = 100 m

a₁ = 100 / (1/2· (2.14 s)² + 2.14 s · 8.26 s)

a₁ = 5.01 m/s²

The initial position of Healan will be the position reached after 3.15 s:

x0₂ = 1/2 · a₂ · (3.15 s)²

And her velocity will be the velocity reached after 3.15 s:

v₂ = a₂ · 3.15 s

Replacing v₂ and x0₂in the equation of position at t = 10.4 s:

1/2 · a₂ · (3.15 s)² + a₂ · 3.15 s · 7.25 s = 100 m

a₂ · (1/2 · (3.15 s)² + 3.15 s · 7.25 s) = 100 m

a₂ = 100 / (1/2 · (3.15 s)² + 3.15 s · 7.25 s)

a₂ = 3.60 m/s²

b)The maximum velocity of  Laura will be:

v₁ = a₁ · 2.14 s

v₁ = 5.01 m/s² · 2.14 s =

v₁ = 10.7 m/s

The maximum velocity of  Healan will be:

v₂ = a · 3.15 s

v₂ = 3.60 m/s² · 3.15 s

v₂ = 11.3 m/s

c) The position of Laura at t = 5.85 s is:

x₁ = x0₁ + v₁ · (5.85 s - 2.14 s)

x₁ = 1/2 · 5.01 m/s² · (2.14 s)² + 10.7 m/s · 3.71 s

x₁ = 51.2 m

The positon of Healan is:

x₂ = x0₂ + v₂ · (5.85 s - 3.15 s)

x₂ = 1/2 · 3.60 m/s² · (3.15 s)²  + 11.3 m/s · 2.7 s

x₂ = 48.4 m

Laura is ahead by (51.2 m - 48.4 m) 2.8 m

d) Let´s find the time at which Healan reaches Laura:

At that time x₁ = x₂  

x₁ = 1/2 · 5.01 m/s² · (2.14 s)² + 10.7 m/s · (t-2.14 s)

x₁ = 11.5 m - 23.0 m + 10.7 m/s · t

x₁ = -11.5 m + 10.7 m/s · t

x₂ = 1/2 · 3.60 m/s² · (3.15 s)² + 11.3 m/s (t - 3.15 s)

x₂ = 17.9 m - 35.6 m + 11.3 m/s · t

x₂ = -17.7 m + 11.3 m/s · t

x₂ = x₁

-17.7 m + 11.3 m/s · t = -11.5 m + 10.7 m/s · t

11.3 m/s · t - 10.7 m/s · t =  -11.5 m + 17.7 m

0.6 m/s · t = 6.2 m

t = 6.2 m/ 0.6 m/s

t = 10.3 s

Healan is behind Laura for 10.3 s

The distance traveled by Healan during that time is:

x₂ = -17.7 m + 11.3 m/s · t

x₂ = -17.7 m + 11.3 m/s · 10.3 s

x₂ = 98.7 m

Healan is behind Laura for 98.7 m.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

gravity

explanation: