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  1. K

    Area under a curve

    but that has nothing to do with the stochastic model or probability model... i need a formula or set of steps not a tool...
  2. K

    Area under a curve

    I have an exponential function graph that was given to me as a question i had to find the points and find three different ways of finding the area under the curve. I found an Empirical model (finding the formula and integrating from one point to another) I found a Finite Element model...
  3. K

    Breaking into traffic at an intersection

    I was given a very open ended question stating... "How Long will it take to break into a line of traffic at a 'T' intersection" I need some sort of start like a theory that includes this kind of thing. We have been studying probability this term so i need something to do with a probabilty...
  4. K

    Integration help

    yer you get (1/30)[(1/-142000)/(142000+30v)] now what?
  5. K

    Integration help

    im only in grade 12.... i dont have the time to take a day off or read calulus texts... i just need some one to tel me what to do and how to integrate it
  6. K

    Integration help

    if this is true: 1/30((142000+30v)/(142000+30v)-142000/(142000+30v)) how do i then integrate it?
  7. K

    Integration help

    nah im only in grade 12 in australia... i just need som help integrating this so i can solve a "flight" plan of a particle with variable acceleration... if you can help me with this it would be great
  8. K

    Integration (displacement or position)

    if you think you can do it without that transition... go for it... that's just how i interpreted the question
  9. K

    Integration help

    1/30((142000+30v)/(142000+30v)-142000/(142000+30v)) like this? what does this achieve?
  10. K

    Integration help

    Homework Statement the trouble is in the integration on the left side (go straight o my answer down the bottom of the page)... please help me... mg-u-kv=m*a where (all constant) m= mass= 3000 g=gravity=10 u=thrust=172000 k=30 v=velocity=variable a=acceleration=variable...
  11. K

    Integration (displacement or position)

    Homework Statement the trouble is in the integration on the left side (go straight o my answer down the bottom of the page)... please help me... mg-u-kv=m*a where (all constant) m= mass= 3000 g=gravity=10 u=thrust=172000 k=30 v=velocity=variable a=acceleration=variable...
  12. K

    Integration (Velocity to Displacement or Position)

    what i ended up with is (e^-t/100)/(-30/100) x e^(-1205.525/-100)/t - 142000/30t that is after integration... is that what you meant?
  13. K

    Integration (Velocity to Displacement or Position)

    i have already tried that... it didnt seem to work... could you show it in latex? so i can see what im doin wrong?
  14. K

    Integration (Velocity to Displacement or Position)

    please help me with the integration in the word document.
  15. K

    Integration (terminal velocity)

    for question 4.) mg-kv^2-u=m*v(dv/ds) i got (using the constant values 1/0.06*ln(|142000+0.03v^2|)=-1/3000*S+c when S=0, v=1000 therefore c=200.921 -1/3000*S=1/0.06*ln(|142000+0.03v^2|)-200.921 when v=0, S=? therefore S=9583.88 m using this same method i need the others...
  16. K

    Integration (terminal velocity)

    1.) mg-kv-u=m*dv/dt left side with respect to dv right side with respect to dt 2.) mg-kv-u=m*v(dv/ds) left side with respect to dv right side with respect to ds 3.) mg-kv^2-u=m*dv/dt left side with respect to dv right side with respect to dt 4.) mg-kv^2-u=m*v(dv/ds)...
  17. K

    Integration (terminal velocity)

    i have 4 equations that need to be integrated 1.) mg-kv-u=m*dv/dt 2.) mg-kv-u=m*v(dv/ds) 3.) mg-kv^2-u=m*dv/dt 4.) mg-kv^2-u=m*v(dv/ds) **if it helps... the values are m=3000 g=10 u= 172000 v changes so leave it as v any help is good help if those are...
  18. K

    Variable acceleration terminal velocity

    thrust=dm/dt x V0=344x5 x 1000 = 172000N. Im talking about a basic scenario here, the model is very basic and alot of assumptions need to be made. K is a constant, m is constant and g is constant. Density at 100km about sea level is negligable and MACH5 is constant no matter what altitude.
  19. K

    Variable acceleration terminal velocity

    you're thinking about it way too literally... the thrust is just saying that there is a force up of 172 000 N im just a year 12 student in australia... dont expect me to now all this stuff lol... all i know is i have to integrate mg-u-kv=m*a -m/k ln (mg-u-kv) = t+c which i think is...
  20. K

    Variable acceleration terminal velocity

    hey thanks for that though... please pass it on to someone that will be able to help me with the algebra thanks again bye
  21. K

    Variable acceleration terminal velocity

    mg is on the y axis downwards but it has a positive value and up is negative... kv and u are both up on the y axis
  22. K

    Variable acceleration terminal velocity

    assume that it is done in the same dimension (same plane) i was given mg-kv-u=m*a where m=mass g=gravity (10) k=drag coefficient u=thrust a=acceleration hope that helps a bit... if that helps i found that when t = 0 v=1000 and therefore i can find k (and i assume that this...
  23. K

    Variable acceleration terminal velocity

    4 conditions - initial velocity = 1000 m/s - height from earth = 100 km - total mass = 3000 kg - rocket capabilities (thrust) = 100 kg/s at MACH 5 i was told that to use viscous and laminar drag as two options and to keep mass constant (to simplify things a little). the space ship is...
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