- #1
Mag|cK
- 36
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Hello, please help me I have a question regarding drag force and Newton's law of motion.
This is quotes from online pdf files from University of Toronto Scarborough.
"the terminal velocity is F/b , where b is drag constant
Let’s solve the equation of motion and see how
this is reflected in the solution. Newton's law
reads
m (dv(t)/dt) = F - b v(t)
which we have written entirely in terms of the
velocity v(t) and its first derivative.
How to solve this? Well, if F were zero, we
would have dv/dt=–(b/m)v, which has its solution
some constant times exp(-bt/m). By inspection,
we find that we can account for nonzero F by
simply adding a constant, F/b. That is,
v(t) = F/b + (constant)exp(-bt/m)
--------------------------------------
Now what i don't understand is the last paragraph, and where the last equation came from. Can anyone explain more clearly? Please help Thank you very much.
This is quotes from online pdf files from University of Toronto Scarborough.
"the terminal velocity is F/b , where b is drag constant
Let’s solve the equation of motion and see how
this is reflected in the solution. Newton's law
reads
m (dv(t)/dt) = F - b v(t)
which we have written entirely in terms of the
velocity v(t) and its first derivative.
How to solve this? Well, if F were zero, we
would have dv/dt=–(b/m)v, which has its solution
some constant times exp(-bt/m). By inspection,
we find that we can account for nonzero F by
simply adding a constant, F/b. That is,
v(t) = F/b + (constant)exp(-bt/m)
--------------------------------------
Now what i don't understand is the last paragraph, and where the last equation came from. Can anyone explain more clearly? Please help Thank you very much.