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...
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...
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
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
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...
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...
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...
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)...
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...
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.
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...
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...
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...