# Search results

1. ### 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. ### 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. ### 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. ### Integration help

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

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

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

15. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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...