Why Doesn't This Work - Easy Kinematics

[erok81]

Homework Statement

An airplane is approaching land with a speed 57 m/s and an angle 15° below the horizontal. The runway is 0.8 km long and the pilot can use the flaps and reverse the engine thrust, to provide a constant deceleration, only after she touches down at the beginning of the runway. How long will it take her to reach the end of the runway and come to a full stop?

Homework Equations

The Attempt at a Solution

My initial veocity is 57cos(15) which leaves me with a landing speed of 55.06m/s since the y-axis velocity cancels..I hope.

I have this solved, but trying two different methods (which should work) in order to verify the answer reveals two completely different answers.

Solving for acceleration:
$$v_f^2 = v_i^2 +2as \rightarrow (v_f^2 - v_i^2)/2s = a$$

Plugging in the numbers I get a = -1.895 m/s^2

Here is where I run into a problem solving for t.

Method One:
$$v_f = v_i +at \rightarrow (v_f - v_i)/a = t$$

Plugging in all of values I have t=29.055s

Method Two:
$$s_f = s_i + v_it + 1/2at^2$$

Here I end up with a quadratic equation with these values:
A= -0.945
B= 55.06
C= 800

This gives me one useable number of 70s.(the other is -12)

If I use -800 for c I come up wtih two close values: 27.67 and 30.59. Now that I am thinking about it. I should use -800 here. But it still doesn't work.

Why is it I can't seem to use the position equation on this one?

Am I missing something that is giving me these two different values? I'd assume both should give me the same value since I am solving for t in both.

 

[erok81]

Final velocity is zero since the plane stops at the end of the runway...at least that is what I am assuming.

 

[Atran]

I am as well a beginner, so I may be wrong too.

Homework Statement

My initial veocity is 57cos(15) which leaves me with a landing speed of 55.06m/s since the y-axis velocity cancels..I hope.

I didn't understand the illustration very well.
If the direction of the velocity 57 m/s is the adjacent, then the 15 angled downward directed velocity is the hypotenuse. Therefore 15 angled downward directed velocity has to be bigger than 57 m/s

Correct me if I'm wrong.

 

[erok81]

I am as well a beginner, so I may be wrong too.



I didn't understand the illustration very well.
If the direction of the velocity 57 m/s is the adjacent, then the 15 angled downward directed velocity is the hypotenuse. Therefore 15 angled downward directed velocity has to be bigger than 57 m/s

Correct me if I'm wrong.

Maybe that is where we went wrong. This is what I have for a diagram...

I am thinking of maybe throwing rolling friction in as well. I don't like this problem because it doesn't say to use friction, like most do, but it doesn't say not to use it either. x2!

 

[Atran]

From what I saw in the picture, I think you're right.

For the time, I used the two methods and they worked equally when I calculated some simple numbers. Therefore I think the false lies somewhere in the calculation.

 

[collinsmark]

Method Two:
$$s_f = s_i + v_it + 1/2at^2$$

Here I end up with a quadratic equation with these values:
A= -0.945
B= 55.06
C= 800

This gives me one useable number of 70s.(the other is -12)

If I use -800 for c I come up wtih two close values: 27.67 and 30.59. Now that I am thinking about it. I should use -800 here. But it still doesn't work.

Your second though was right. You need to use -800 for C.

Try redoing what you have done so far (starting by recalculating the acceleration), and pay extra attention to the precision. You'll find an interesting thing happens with the
B² - 4AC
part of the equation. But you can't ignore precision. Then you'll find that method 2 gives you the same answer as method 1.

 

[ideasrule]

My initial veocity is 57cos(15) which leaves me with a landing speed of 55.06m/s since the y-axis velocity cancels..I hope.

For this question, yes, that's how you're supposed to do it. In reality, the pilot brings the plane's nose up just before reaching the runway, then flies parallel to the runway. The plane slows down until the lift it generates is less than the plane's weight, at which plane it falls the remaining 30 cm or so onto the ground.

Method Two:
$$s_f = s_i + v_it + 1/2at^2$$

What did you plug in for the numbers? Was it a=-1.895 m/s, s_f=800, a=-1.895 m/s^2, s_i=0, and v_i=55 m/s?

 

[Atran]

collinsmark,

There are two possible answers, one of them matches erok81's first method answer,
but what the second possible answer stands for (t = 30.59... s)?

 

[ideasrule]

There is only one possible answer if the quadratic equation is solved correctly.

 

[collinsmark]

collinsmark,

There are two possible answers, one of them matches erok81's first method answer,
but what the second possible answer stands for (t = 30.59... s)?

The precision is important. For example, in your earlier calculations, you had A = -0.945 (even without the extra precision, it should have been -0.947). Recalculate that, starting over, with something more along the lines of -0.947376 or so.

 

[erok81]

For this question, yes, that's how you're supposed to do it. In reality, the pilot brings the plane's nose up just before reaching the runway, then flies parallel to the runway. The plane slows down until the lift it generates is less than the plane's weight, at which plane it falls the remaining 30 cm or so onto the ground.



What did you plug in for the numbers? Was it a=-1.895 m/s, s_f=800, a=-1.895 m/s^2, s_i=0, and v_i=55 m/s?

Yeah, those were the numbers. Excect v_i was 55.06.

The precision is important. For example, in your earlier calculations, you had A = -0.945 (even without the extra precision, it should have been -0.947). Recalculate that, starting over, with something more along the lines of -0.947376 or so.

Hmm...

Now using these numbers:
A= -0.947376
B= 55.06
C= -800

I get no reals.

 

[erok81]

Oh lovely. That's from my calculator.

If I try an online quadratic calc, there is only one solution.

I wonder why that is.

 

[erok81]

Well besides my pos calculator, it's solved! Thanks for the help everyone. It is much appreciated.

Now I can see why the numbers had to be very exact. The solutions were the same.