Plane Landing on Tropical Island: Max Acceleration

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A plane landing on a small tropical island with a 66 m runway and an initial speed of 58 m/s requires calculating maximum deceleration using the formula v^2 = Vo^2 + 2a(x-Xo). The final velocity (Vf) is zero upon landing, leading to the rearranged equation a = (-Vi^2)/(2*d), resulting in a maximum deceleration of -25.5 m/s^2. Discussions also included calculating the time to stop and the distance traveled during acceleration and deceleration phases of a car's motion. The correct approach emphasizes adding distances from both phases rather than subtracting them, as they occur in the same directional vector.
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A plane landing on a small tropical island has
just 66 m of runway on which to stop.
If its intial speed is 58 m/s, what is the
maximum acceleration of the plane during
landing, assuming it to be constant? Answer
in units of m/s^2.

i know I am supposed to use the equation v^2 = Vo^2 + 2a(x-Xo) but wht is V^2
 
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What formula would you have to use??

Daniel.
 
dextercioby said:
What formula would you have to use??

Daniel.


v^2 = Vo^2 + 2a(x-Xo) but wht is V^2
 
Well,the plane has to stop,so the final velocity should be...??

Daniel.
 
dextercioby said:
Well,the plane has to stop,so the final velocity should be...??

Daniel.


ah ha, ok i understand
 
follow up question

How long does it take for the plane to stop
with this acceleration? Answer in units of s.

i used the equation X = Xo + Vot + 1/2at^2 is this correct?
 
How about
v_{fin}=v_{init}+at

?? :wink:

The one u posted is correct as well,but te one suggested by me is more direct.

Daniel.
 
dextercioby said:
How about
v_{fin}=v_{init}+at

?? :wink:

The one u posted is correct as well,but te one suggested by me is more direct.

Daniel.


i just got that
 
one more question

A car starts from rest and travels for 6.4 s with
a uniform acceleration of 3.3 m/s^2. The driver
then applies the brakes, causing a uniform
acceleration of 1.5 m/s^2.
If the brakes are applied for 2.5 s, how fast
is the car going at the end of the braking
period? Answer in units of m/s.

this is what I did, i subtracted the final acceleration from the initial and the the time for braking from the total time the car was traveling. then i used the equation V = Vo + at is that what you would do?
 
  • #10
No,i would compute the velocity after 6.4 secs and then the final velocity.Do the same & it'll be okay.

Daniel.
 
  • #11
one final add on

same question, How far has it gone? Answer in units of m.

i used the formula X = Xo + Vot + 1/2at^2 for each time and acceleration then i subtracted the distance i got from the deceleration from the the distance i got for the acceleration
 
  • #12
Why did u subtract them...?It doesn't go backwards... :rolleyes: :wink:

Daniel.
 
  • #13
dextercioby said:
Why did u subtract them...?It doesn't go backwards... :rolleyes: :wink:

Daniel.

because one answer was negative even though distance cannot be negative( the decleration distance) and the acceleration distance was positive. am i supposed to add them?
 
  • #14
Of course,it decelerates in the same direction as it was accelerating (the velocity vector keeps direction,just the modulus varies),so they have to be added.

Daniel.
 
  • #15
dextercioby said:
Of course,it decelerates in the same direction as it was accelerating (the velocity vector keeps direction,just the modulus varies),so they have to be added.

Daniel.


im getting 72.27meters, can u check that for me? i think its wrong
 
Last edited:
  • #16
Nope,should be roughly 110m...


Daniel.
 
  • #17
plane on island

Vf^2 = Vi^2 + 2ad

Vf = 0. because the plane will be at stopped after it lands

rearrange formula

-Vi^ = 2ad

a = (-Vi^2)/(2*d)

a = (-58^2)/(2*66)

a = -25.5 m/s^2

aceleration is negative because the plane is slowing down while it lands.
 
  • #18
v2 is zero, once the plane lands it dosent have a velocity now does it
 
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