One Dimensional Kinematics Jet Problem

In summary: X-Xo = Vo*t + (1/2)*a*t^2.Also, you are applying the quadratic formula incorrectly. You are divining by (2*constant) where you should be doing (2*a), where a is the coefficient of the squared term of the variable which is t^2 in our case.
  • #1
Fumbalodian
5
0

Homework Statement



A Boeing 747 "Jumbo Jet" has a length of 59.7 m. The runway on which the plane lands intersects another runway. The width of the intersection is 29.9 m. The plane decelerates through the intersection at a rate of 6.33 m/s2 and clears it with a final speed of 42.3 m/s. How much time is needed for the plane to clear the intersection?



Homework Equations



V^2 = V0^2 + 2a (X - X0)

X = 1/2 (V0 + V) t



The Attempt at a Solution



Knowns:
X0 = 0m
X = 89.6m
V0 = Unknown
V = 42.3 m/s
a = -6.33 m/s^2
t = Unknown

I began by trying to solve for the initial velocity of the jet using the first of the above equations. I am getting 95.7 m/s but I am not sure if I am simply doing the algebra incorrectly or something else. It would help to get step by step help on the algebra as well and not skip any steps.

I then used 95.7 m/s in the second equation to solve for time. For this I am getting
1.3s but the homework webpage says that isn't correct.

I have tried repeatedly to work this through and I just can't seem to get it.

Thank you
 
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  • #2
Do a trick here. Time reverse the situation. Imagine the jet is moving backward with the same acceleration (not deceleration) of +6.33 m/s^2. It's initial velo is what is given as the final velo of +42.3 m/s. Then you'll get the answer in one shot.

X = Vi*t + (1/2)at^2.

You can do this because the time for going from Vi to Vf with deceleration a is the same for going from Vf to Vi with accn a.
 
  • #3
Fumbalodian said:

Homework Equations



V^2 = V0^2 + 2a (X - X0)

X = 1/2 (V0 + V) t



The Attempt at a Solution



Knowns:
X0 = 0m
X = 89.6m
V0 = Unknown
V = 42.3 m/s
a = -6.33 m/s^2
t = Unknown

I began by trying to solve for the initial velocity of the jet using the first of the above equations. I am getting 95.7 m/s but I am not sure if I am simply doing the algebra incorrectly or something else.


You're doing the algebra incorrectly. Can you show your steps, and we'll see where you're going wrong?
 
Last edited:
  • #4
Shooting star said:
Do a trick here. Time reverse the situation. Imagine the jet is moving backward with the same acceleration (not deceleration) of +6.33 m/s^2. It's initial velo is what is given as the final velo of +42.3 m/s. Then you'll get the answer in one shot.

X = Vi*t + (1/2)at^2.

You can do this because the time for going from Vi to Vf with deceleration a is the same for going from Vf to Vi with accn a.

I tried plugging in the numbers to that formula but I can't seem to get the right answer.

I did it as follows.

89.6m = 42.3 m/s*t + (1/2) 6.33 m/s^2*t^2

First off, was that what you meant?

I tried to solve it using the quadratic forumula but it doesn't seem to be working out.
0= -89.6m + 42.3m/s*t +(1/2) 6.33m/s^2*t^2

t= ((-42.3 + √(42.3^2-4(-89.6)(3.17))) / (2(-89.6)))

What might I be doing wrong?
 
  • #5
I meant exactly that.

I got 5.89 s. Find the positive root of the quadratic eqn in 't'. A matter of one minute!
 
  • #6
Shooting star said:
I meant exactly that.

I got 5.89 s. Find the positive root of the quadratic eqn in 't'. A matter of one minute!

Hmmm...

I tried checking that answer on the WileyPlus webpage we use and it says 5.89s is not a correct answer.
 
  • #7
Maybe I did some arithmetical error. You don't know how to solve quad eqn? Why do you have to go to some web page?

I did it again. I got 0.93 s.
 
  • #8
Shooting star said:
Maybe I did some arithmetical error. You don't know how to solve quad eqn? Why do you have to go to some web page?

I did it again. I got 0.93 s.


I got those answers too. The webpage is how my teacher assigns us homework which is why I am using that. It doesn't like 0.93 s either. I guess I give up. Thanks for your help though.
 
  • #9
V^2 = V0^2 + 2a (X - X0)
42.3^2 = Vo^2 - 2*6.33*89.6
Vo = 54.1m/s
V = Vo + at or 42.3 = 54.1 - 6.33t
t = (54.1-42.3)/6.33 = 1.86s
 
Last edited:
  • #10
The equation you used first were the easy way to start. if you follow your steps again you should find that the initial velocity is 54.07m/s.


for the quadratic equation you need to used the initial speed not the final.

X-Xo = Vo*t + (1/2)*a*t^2.

also you are applying the quadratic formula incorrectly. you are divining by (2*constant) were you should be doing (2*a), where a is the coefficient of the squared term of the variable which is t^2 in our case
 
  • #11
All right, I divided by an extra 2 -- the 2a in the denominator of the qudratic roots. So, 2*0.93 = 1.86.
 
  • #12
robphy said:
Why is "X = 89.6m"?
How far does the plane travel as it passes through the intersection?

Width + length of plane.
 
  • #13
Shooting star said:
Width + length of plane.

I saw it after posting... but didn't delete it quick enough. :rolleyes:
 
  • #14
Fumbalodian said:
I got those answers too. The webpage is how my teacher assigns us homework which is why I am using that. It doesn't like 0.93 s either. I guess I give up. Thanks for your help though.

See if the web page likes the correct answer now. And never give up.
 
  • #15
Shooting star said:
See if the web page likes the correct answer now. And never give up.

Thanks for the help. That answer is correct.
 

1. What is "One Dimensional Kinematics Jet Problem"?

"One Dimensional Kinematics Jet Problem" is a type of physics problem that involves analyzing the motion of an object in one-dimensional space, specifically in the context of a jet or aircraft. It involves using equations and principles of kinematics to determine the position, velocity, and acceleration of the object at different points in time.

2. How do you solve a "One Dimensional Kinematics Jet Problem"?

To solve a "One Dimensional Kinematics Jet Problem", you first need to identify the given information, such as initial and final positions, velocities, and accelerations. Then, you can use equations such as the kinematic equations and the equation of motion to solve for the unknown variables. It is important to carefully label and convert units in order to obtain accurate results.

3. What are some common equations used in "One Dimensional Kinematics Jet Problems"?

Some common equations used in "One Dimensional Kinematics Jet Problems" include the kinematic equations, which relate displacement, velocity, acceleration, and time, as well as the equation of motion, which relates force, mass, acceleration, and displacement. The equations of motion for constant acceleration are also frequently used.

4. What are some real-life applications of "One Dimensional Kinematics Jet Problems"?

"One Dimensional Kinematics Jet Problems" can be applied to real-life situations such as determining the takeoff and landing distances for aircraft, calculating the maximum height and range of a projectile, and predicting the motion of a roller coaster. It is also used in the design and optimization of transportation systems and in studying the effects of air resistance on objects in motion.

5. What are some common mistakes to avoid when solving "One Dimensional Kinematics Jet Problems"?

Some common mistakes to avoid when solving "One Dimensional Kinematics Jet Problems" include using the wrong equation or formula, forgetting to convert units, and not paying attention to the direction of motion. It is also important to carefully consider the given information and make sure it is consistent with the equations being used. Additionally, double-checking your calculations and using significant figures can help avoid errors.

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