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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

Respuesta :

mauricioalessandrell

Answer:

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.