Basic linear motion question.Is it good?

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SUMMARY

This discussion focuses on calculating the minimum runway length required for airplanes based on their acceleration and take-off speeds. For an airplane with a lowest acceleration rate of 3 m/s² and a take-off speed of 65 m/s, the minimum runway length is determined to be 704.2 meters. Another scenario involves an airplane accelerating at 3.2 m/s² for 32.8 seconds, resulting in a distance of 1721.3 meters before take-off. Additionally, an airplane requiring 1365 meters to reach a take-off speed of 88.3 m/s has an acceleration of 2.856 m/s² and a time to reach this speed of 30.9 seconds.

PREREQUISITES
  • Understanding of kinematic equations: v = u + at, v² = u² + 2as, s = ut + (1/2)at²
  • Basic knowledge of physics concepts related to motion and acceleration
  • Familiarity with units of measurement in physics (meters, seconds)
  • Ability to perform algebraic manipulations for solving equations
NEXT STEPS
  • Study advanced kinematic equations for varying acceleration scenarios
  • Explore real-world applications of physics in aviation and runway design
  • Learn about the impact of different acceleration rates on aircraft performance
  • Investigate safety regulations and standards for runway lengths in aviation
USEFUL FOR

Aerospace engineers, physics students, aviation safety analysts, and anyone involved in airport design and aircraft performance optimization would benefit from this discussion.

german1234
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Homework Statement


Airport runway designing:

(i)lowest acceleration rate of any plane is likely to be 3m/s^2.Take off speed for this airplane will be 65m/s.what is minimum allowed length of runway?

(ii)If airplane accelerates down runway at 3.2m/s^ for 32.8 seconds until it lifts off..determine distance traveled before take-off.

(iii)Another airplane has take off speed of 88.3m/s and needs 1365m to reach this speed.determine acceleration of plane and time taken to reach this speed.

Homework Equations


v=u+at

v^2=u^2+2as

s=ut+(1/2)at^2


The Attempt at a Solution




(i) a=3
u=0
v=65
s=? v^2=u^2+2as

65^2=0^2+2(3)s
4225=6s
s=4225/3
s= 704.2 m

(ii) a=3.2
t=32.8
u=0
s=?
s=ut+(1/2)at^2

s=0(32.8)+(1/2)(3.2)(32.8)^2

s=1721.3m

(iii)v=88.3
u=0
s=1365
t=?

v^2=u^2+2as

(88.3)^2=0^2+2a(1365)

7796.89=2730a

a=2.856m/s^2

Now:
v=u+at

88.3=0+2.856(t)

t=30.9 seconds

Thanks in advance
 
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