One Dimensional Kinematics Jet Problem

[Fumbalodian]

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

 

[Shooting Star]

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.

 

[dotman]

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?

 

[Fumbalodian]

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?

 

[Shooting Star]

I meant exactly that.

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

 

[Fumbalodian]

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.

 

[Shooting Star]

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.

 

[Fumbalodian]

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.

 

[rl.bhat]

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

 

[shehab]

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

 

[Shooting Star]

All right, I divided by an extra 2 -- the 2a in the denominator of the qudratic roots. So, 2*0.93 = 1.86.

 

[Shooting Star]

Why is "X = 89.6m"?
How far does the plane travel as it passes through the intersection?

Width + length of plane.

 

[robphy]

Width + length of plane.

I saw it after posting... but didn't delete it quick enough.

 

[Shooting Star]

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.

 

[Fumbalodian]

See if the web page likes the correct answer now. And never give up.

Thanks for the help. That answer is correct.