Calculating Time for an Airplane to Come to Rest Using Classical Mechanics

AI Thread Summary
An airplane landing at 100 m/s travels 1000 meters before coming to rest, prompting a calculation of the time taken to stop. The initial attempt used the equation Xf = Xi + Vi*t, yielding an incorrect time of 10 seconds. The correct approach involves recognizing the need for deceleration and applying the mean velocity concept, resulting in a time of 20 seconds. The discussion emphasizes the importance of using appropriate equations for constant acceleration, such as Xf = Vi*t + 1/2 a*t^2. Ultimately, the correct answer is derived from understanding the principles of kinematics and energy conservation.
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Homework Statement



An airplane touches down at a speed of 100m/s. It travels 1000 metres along the runway
while deceleration at a constant rate before coming to rest. How long did it take the airplane to come to rest on the runway?

Xi=0m/s Xf=1000m Vix=100m/s

Homework Equations


Xf=Xi+Vixt

The Attempt at a Solution


1000=0+100t
t=10s

Why is the answer 20 s? and why does this formula fail?
 
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It seems that your particle isn't affected by any forces, thus it keeps on moving in the same direction with constant speed.
 
the plane need decelation the speed from 100 m/s to 0 m/s so the mean velocity is 50 m/s then t = 20 s.
 
Your math is correct, but your equation is wrong. I've calculated it and the answer
t=20s is correct. Try revisiting your equations.
 
No, it's not wrong at all. you could use the equation and mean velocity. But if you need to use the standard equation. it will be
Xf = Vi*t + 1/2 a*t^2 (1)
Vf = Vi + a*t (2)
a = (Vf - Vi) /t (3)
Xf = Vi*t + 1/2 (Vf -Vi) * t
Xf = 1/2 (Vf +Vi)*t (4)
Xf = 1000 m , Vi = 100 m/s , Vf = 0 m/s
 
Isn't this just constant acceleration motion? Just use the constant acceleration formulae to find the acceleration first.

Useful formulae:
v2 - vi2 = 2aΔx
v=vi + at
 
yes, you are right.
You can use the simple form of energy conservation
1/2 m vf2 - 1/2 m vi2 = m a Δx
=> v2 - vi2 = 2aΔx
Thanks.
 
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