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)
<u>a₁ = 5.01 m/s²</u>
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)
<u>a₂ = 3.60 m/s²</u>
b)The maximum velocity of Laura will be:
v₁ = a₁ · 2.14 s
v₁ = 5.01 m/s² · 2.14 s =
<u>v₁ = 10.7 m/s</u>
The maximum velocity of Healan will be:
v₂ = a · 3.15 s
v₂ = 3.60 m/s² · 3.15 s
<u>v₂ = 11.3 m/s</u>
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
<u>x₁ = 51.2 m</u>
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
<u>x₂ = 48.4 m</u>
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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
<u>t = 10.3 s </u>
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
<u>x₂ = 98.7 m</u>
Healan is behind Laura for 98.7 m.
