Can Someone quickly help me !!! (1 Viewer)

[liveit_up]

Q) where a,v, and x are the respective instantanoues acceleration, velocity nd postion of a particle moving on a straight line at time t.

i) if v=-kv, find x in terms of t

solution

(dx/dt)=-kx

dx/x=-kx dt

taking intergrals of both sides gives

lnx=-kt+ln A <----------------------- ????? Why is the A there

therefore x=Ae^(-kt) where did it come ????


(we have just started mechanics arrggg !!!!)

 

[Affinity]

(1) it's not v=-kv it's v = -kx

(2) the 2nd line of the solution should be dx/x = -k dt (without the x)

(3) the ln(A) is the constant of integration. like +C.

(4) Go back and master 3 unit application of calculus, this particular question is 3 U.

 

[liveit_up]

just one last question it might seem stupid
how do u get the constant to equal = ln(A)

 

[toisthbe]

in the terry lee book, i believe that ln(A) is an assumed constant, thats one bad thing bout that book.

 

[liveit_up]

then in part iii) the constant is not assummed to be ln(A)

why?????

 

[Pythagoras]

I can completely understand your confusion with this point. Its basically just mathematicians being their usual elegant selves. I suspect also that they thought this up later after they already had the answer to the integration.

Try doing the integration in the usual way with c as the constant of integration. You end up with:

x = e^(kc-kt)
= e^kc x e^-kt

e^kc is just a constant which we may as well call A.

Therefore, x = Ae^-kt

If we now work backwards and take natural logs of both sides of this equation, we get

lnx = lnA +-kt

So when you compare this line with your earlier work you can see that the constant of integration is ln A.

It is quite legal to call the constant of integration, lnA. It looks important but really, A is just another name for a constant and its natural log is also a constant. Mathematicians do this because they know where the mathematics is heading. Its smarter and more efficient to call the constant lnA so that you can jump neatly to x =e^(lnA-kt)
so x = Ae^-kt (using the result e^lnx = x)