# 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

## Homework Equations

accel=v*dv/ds=dv/dt=a

## The Attempt at a Solution

30000-172000-30v = 3000*v*dv/ds
-142000-30v = 3000*v*dv/ds
-(142000+30v) = 3000*v*dv/ds
142000+30v = -3000*v*dv/ds
(142000+30v)/v*dv = -3000*ds
v/(142000+30v)*dv = -1/3000*ds

and then integrate both sides, this is where my calculus stops

i know the right side (-1/3000*ds) becomes -1/3000s + c
where
s= displacement or position and
c= constant
i have 2 scenarios

when s=0, v=1000
and what i need to find is when v=0 what is s?

so once i integrate the left side i sub those values in.

Meir Achuz
Homework Helper
Gold Member
Your integrand is of the form x/(ax+b).
Write it as (1/a)[(ax+b)/(ax+b)-b/(ax+b)].

1/30((142000+30v)/(142000+30v)-142000/(142000+30v))

like this?

what does this achieve?

malawi_glenn
Homework Helper
1/30((142000+30v)/(142000+30v)-142000/(142000+30v))

like this?

what does this achieve?

Well integrating functions of the type:

x/(b-x) is pretty hard if you dont rewrite them..

You will after a rewriting get things that has ln(x)-functions as primitive functions.

Have you taken courses in calculus?

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

if this is true:

1/30((142000+30v)/(142000+30v)-142000/(142000+30v))

how do i then integrate it?

Meir Achuz
Homework Helper
Gold Member
If you can't integrate that, take a day off and read a calculus text before continuing physics.

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

Meir Achuz
Homework Helper
Gold Member
Can you divide (142000+30v) by (142000+30v)?

1/30((142000+30v)/(142000+30v)-142000/(142000+30v))

like this?

what does this achieve?

Why do you insist on sticking those ugly numbers in, at every opportunity? Try working with the constants/variables you defined; much easier to solve problems this way (and makes troubleshooting a breeze.)

yer you get

(1/30)[(1/-142000)/(142000+30v)]

now what?