Calculating Jet Velocity with Kinematic Equations

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
11 replies · 2K views
deaninator
Messages
64
Reaction score
0

Homework Statement


Mike is piloting a lear jet traveling at 203.6 m/s over a distance of 2395 meters. How fast will the jet be moving at the end of this acceleration?

Homework Equations


V = D/T?
You need to incorporate one of the "kinematic equations" in order to solve.
V=Vo + at
X=1/2(Vo -V)T
X=volt + 1/2at2
V2=Vo2 + 2ax

(If a 2 is after a letter then it means squared)
X=displacement
A=Acceleration
T=Tome
Vo=Initial Velocity
V=Velocity

The Attempt at a Solution



V2 = 203.6 + 2(15.2)(2395)
V2 = 270.21...That answer is NOT correct.
 
Last edited:
Physics news on Phys.org
The question says "a lear jet traveling at 203.6 m/s over a distance of 2395 meters". There is no acceleration implied there, you are given a speed not acceleration, and so the jet would be traveling at 203.6m/s at the end of this distance. Or is there more to this question?

Also, please show an attempt at the solution.

Jared
 
jarednjames said:
The question says "a lear jet traveling at 203.6 m/s over a distance of 2395 meters". There is no acceleration implied there, you are given a speed not acceleration, and so the jet would be traveling at 203.6m/s at the end of this distance. Or is there more to this question?

Also, please show an attempt at the solution.

Jared

I'm sorry, the acceleration is 15.2 m/s/s
 
Much better.

But I still need an attempt at a solution.

Jared
 
jarednjames said:
The question says "a lear jet traveling at 203.6 m/s over a distance of 2395 meters". There is no acceleration implied there, you are given a speed not acceleration, and so the jet would be traveling at 203.6m/s at the end of this distance. Or is there more to this question?

Also, please show an attempt at the solution.

Jared

jarednjames said:
Much better.

But I still need an attempt at a solution.

Jared

Sorry I forgot that, it is now up.
 
Using V2=Vo2 + 2ax

You know Vo, a and x.

Plug in your values and out comes your answer for final velocity V.

Note that in your solution attempt you didn't square V and Vo so your answer is wrong.

Once you get a value for V2 you need to square root it to get V.

Jared
 
jarednjames said:
Using V2=Vo2 + 2ax

You know Vo, a and x.

Plug in your values and out comes your answer for final velocity V.

Jared

Yes, I did that, and the product was 270.21...that is unfortunately not correct according to my computer.
 
No, you didn't do that.

Refresh the page. I updated what you did wrong.

Jared
 
jarednjames said:
No, you didn't do that.

Refresh the page. I updated what you did wrong.

Jared

All right. V2 = 0 + 2(15.2)(2395)
V2 = 73011.6
V = 270.206 M/S
 
deaninator said:
All right. V2 = 0 + 2(15.2)(2395)
V2 = 73011.6
V = 270.206 M/S

Wait was the initial velocity 0 or 203.5 m/s?
 
Vo = 203.6, x = 2395 and a = 15.2

V2 = 203.62 + (2*15.2*2395)

V2 = 41452.96 + 72808 = 114260.96

V = Square Root (114260.96)

Jared
 
jarednjames said:
Vo = 203.6, x = 2395 and a = 15.2

V2 = 203.62 + (2*15.2*2395)

V2 = 41452.96 + 72808 = 114260.96

V = Square Root (114260.96)

Jared

I ended up with 338.03 m/s and that was correct.
Thank you sir.