Distance traveled during speeding-up phase of a propelled sled

[IAmPat]

Homework Statement

At the Aircraft Landing Dynamics Facility located at NASA's Langley Research Center in Virginia, a water-jet nozzle propels a 46,000 kg sled from zero to 400 km/hour in 2.5 seconds. Assuming a constant acceleration for the sled, what distance does it travel during the speeding-up phase (meters)?

Homework Equations

The Attempt at a Solution

400km/h = 400,000 m/h
3600 seconds in 1 hour
400,000 * 3600 = 1,440,000,000 m/s = Final Velocity

Vo = 0 m/s
Vf = 1,440,000,000 m/s
t = 2.5s

With a constant acceleration that would mean

a = 1,440,000,000 / 2.5 = 576,000,000 m/s^2

but if I multiply the acceleration by 2.5 I end up with my original final velocity, which doesn't make sense... Am I missing a force or something else? This seems like it should be an easy problem, but for some reason I'm just blanking on it.

 

[cupid.callin]

Hi PAT

400km/h = 400,000 m/h
3600 seconds in 1 hour
400,000 * 3600 = 1,440,000,000 m/s = Final Velocity

Do it again
Its wrong!

And didn't you notice its greater than speed of light ... How is that possible

 

[IAmPat]

Hi PAT



Do it again
Its wrong!

And didn't you notice its greater than speed of light ... How is that possible

This is what I get for working after being completely mentally drained.

Ok so, it should have been 400,000 DIVIDED by 3600

which gives me 111.1111~ m/s

-> a = 111.111~ / 2.5
-> a = 44.4444~

So now I have my (correct?) acceleration. But once again I'm stuck. If I multiply the acceleration by the time it was speeding up, I get the original 111.1111~. Is that right?

 

[cupid.callin]

which gives me 111.1111~ m/s

-> a = 111.111~ / 2.5
-> a = 44.4444~

Yes that looks correct

So now I have my (correct?) acceleration. But once again I'm stuck. If I multiply the acceleration by the time it was speeding up, I get the original 111.1111~. Is that right?

Yes of course that will happen. Its according to Newton's first eqⁿ of motion
You should be more attentive in your lectures :rofl:

 

[IAmPat]

Yes that looks correct



Yes of course that will happen. Its according to Newton's first eqⁿ of motion
You should be more attentive in your lectures :rofl:

The only reason I'm confused as to if it's correct or not is because when I enter in "111.111" as my answer, the grading system is marking it wrong.

 

[cupid.callin]

The only reason I'm confused as to if it's correct or not is because when I enter in "111.111" as my answer, the grading system is marking it wrong.

Of course it will say its wrong ... its final speed not velocity :uhh:

use second eqn of motion to find distance

s = ut + 0.5 at²

Why you came up with s = at ?

 

[Tim D.]

Make sure to always draw a diagram and a fbd (free body diagram) it will always work! ;)

v=v(o)+at
x=x(0)+v(0)t+at^2
v^2=v(0)^2+2a(x-x(0))
x-x(0)=((v+v(0))/2)t
*recall your constant accel equations?

Then find your "y stuff" and then find your "x stuff"
t (time) goes both ways!

also remember the
v(x)= v + sin (theta)
v(y)= v + cos (theta)

 

[Tim D.]

Also make your x,y axis as simple as possible. Try to put the axis along the most "kinematics" or "forces" what ever you working with. Its usually best to put a(y) "gravity" on the positive y axis.

 

[cupid.callin]

Make sure to always draw a diagram and a fbd (free body diagram) it will always work! ;)

v=v(o)+at
x=x(0)+v(0)t+at^2
v^2=v(0)^2+2a(x-x(0))
x-x(0)=((v+v(0))/2)t
*recall your constant accel equations?

Then find your "y stuff" and then find your "x stuff"
t (time) goes both ways!

also remember the
v(x)= v + sin (theta)
v(y)= v + cos (theta)

Please don't give full solutions
Let the OP do the problem himself

 

[Tim D.]

LOL, that's not full solutions, that's just the equations which he should already know!