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BlueSky06

This question s is really asking us to first find the length of runway needed to achieve takeoff speed. Once we find this minimum length, we’ll multiple it by three and compare it to a 2.5 mile runway. The simplest way to solve for this minimum length is to use the answers to the last questions. We will plug acceleration we found and the minimum time required into

 

[tex] x_{f} = \frac{1} {2} a * [/tex]Δ[tex]t^2 = \frac{1} {2} * 2.3 \frac{m} {s^2} * (34.8s)^2 = 1,393m = 1.39 km[/tex]

 

And we find the minimum runway length is 1.39 km or 0.863 miles (using the conversions in the inside cover of the text). The safe runway length is then 2.59 miles. So, the 2.5 mile long runway in the question is too short, by a smidgen, to launch this jetliner safely

johanrusli

The minimum length runway this aircraft can use is about 4.1 km

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

Let's look at the table in the attachment

Given:

u = 0 m/s

v = 79 m/s

a = 23 / 10 = 2.3 m/s²

Unknown:

d = ?

Solution:

Let's calculate the takeoff distance.

[tex]v^2 = u^2 + 2as[/tex]

[tex]79^2 = 0^2 + 2(2.3)s[/tex]

[tex]6241 = 4.6s[/tex]

[tex]s = 6241 \div 4.6[/tex]

[tex]s \approx 1400 ~ m[/tex]

The length of the runway must be three times the takeoff distance.

[tex]d = 3s[/tex]

[tex]d = 3(6241 \div 4.6)[/tex]

[tex]d \approx 4100 ~ m[/tex]

Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

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