Where is my mistake?

  • Thread starter omagdon7
  • Start date
  • #1
95
0
The Warrior can take off when its airspeed (speed of the air flowing over the wing) is equal to 100 km/hr. What is the length of runway, in meters, required for the plane to take off if there is a 12 km/hr head wind? Note: the shortest runway at the Rochester International Airport has a length of 1000 m. Take the acceleration from the the answer of the previous question. The acceleration from past problem was .444444444444444m/s^2

Okay so first I set
100=-12+.444444444t
then 112=.444444444t
convert 112 to m/s
31.1111111=.4444444444t
t=70
then I use that time in
x=-3.333t+(1/2)(.44444444)t^2
and solve for x right? and I get
855.57m as the length of the runway but this is incorrect so where exactly is my mistake?
 

Answers and Replies

  • #2
BobG
Science Advisor
Homework Helper
185
82
A head wind means the wind is hitting the windshield (and the front of the wings). A tail wind means the wind is hitting the tail (and the back of the wings).

In other words, while sitting still, the plane has an airspeed of 12 km/hr (not -12 km/hr)
 
  • #3
95
0
so then...
100=12+.44444t
88=.444444t
88/3.6/.4444444=t
t=55
so
x=3.33333(55)+(1/2)(.444444)(55)^2
and
x=855.5
lol same answer and it is still wrong so what is my mistake now.
 
  • #4
95
0
Okay nm I figured it out the wind speed was only providing lift it was not acting on the plane as if the plane was a particle so after solving for T=55 I simply needed to plug it into x=(1/2)(.44444)t^2.

That was tricky
 

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