- #36
Ashley1nOnly
- 132
- 3
what's the purpose of switching the two values?
first integral
-∫ 1/(v' -vter) dv' second integral
∫ 1/(v'+ vter) dv'
first integral
-∫ 1/(v' -vter) dv' second integral
∫ 1/(v'+ vter) dv'
Switch what two values?Ashley1nOnly said:what's the purpose of switching the two values?
first integral
-∫ 1/(v' -vter) dv'second integral
∫ 1/(v'+ vter) dv'
Charles Link said:The mistake you originally made, and otherwise you had the calculation correct, was to integrate ## \frac{1}{vter-v' } ##. In this case, ## v' ## is the variable of integration, and it needs to be in the form ## \int \frac{dx}{x+a}=ln|x+a| ##. When you have it in the form ## \int \frac{dx}{a-x} ##, it is not in a standard form where you can use the result ## \ln|a-x| ##. That is incorrect. It is ## -\ln|a-x| ##. ## \\ ## Just because ## \int \frac{dx}{x+a}=\ln|x+a| ##, that does not mean ## \int \frac{dx}{a-x}=\ln|a-x| ##.
Technically you have two errors here, which cancel (by coincidence).,Ashley1nOnly said:first integral
-∫ 1/(v' -vter) dv'
-ln(v'-vter) } with limit from 0 to V
-ln(v-vter) +ln(0-vter)
-ln(v-vter)+ln(-vter)
-ln(v-vter / -vter)
-ln(-v/vter +1)
...
Yes, if you are referring to the 'first integral' and 'second integral' of your post (#28) .Ashley1nOnly said:so when you integrate the first one you get
-ln|vt-v|
and the second one
ln|vt+v|
without applying the limits
Fine.Ashley1nOnly said:applying limits
first integral
-ln|vt -v| - [-ln |vt-0|]
-ln|vt -v| +ln|vt|
ln| vt/ (vt-v) |
ln| (1-vt/v)|
second integral
ln|vt+v| -ln|vt+0|
ln|vt+v|-ln|vt|
ln| (vt + v )/(vt) |
ln| (1+ v/vt)
−ln|vt −v| + ln|vt| + ln|vt+v| − ln|vt| =ln|vt+v|−ln|vt −v|=ln[(vt+v)/(vt-v)].SammyS said:−ln|vt −v| + ln|vt| + ln|vt+v| − ln|vt|
Out of post 44, this step is incorrect. It is also really an unnecessary step, but if you do want to proceed that way, you do need to get the algebra correct.Ashley1nOnly said:ln| vt/ (vt-v) |
ln| (1-vt/v)|
Please read posts (#46) and (#49), and save yourself some work.Ashley1nOnly said:Right I totally skipped over that. I can't divide like that.
first integral
ln[ vt/ (vt-v)]
second integral
ln[ 1 + v/vt]
ln[ vt/ (vt-v)]+ln[ 1 + v/vt]
=ln [ (vt+v )/ (vt-v)]
Look at the OP (original post):Ashley1nOnly said:How do I calculate the F?
Looks like you might have gone back to your hyperbolic tangent =tanh solution, but as ## t \rightarrow +\infty ## in the numerator and denominator, the limit will be 1 for that fraction and not zero. So this answer is incorrect. ## \\ ## From post 53, you should also be able to get the correct answer very quickly by looking at the denominator ## v_T-v ## of the natural log function argument there. ## \\ ## Here's a hint: Suppose we let ## v \rightarrow v_T ##, so that ## v_T-v \approx 0 ##, what happens to that natural log function?Ashley1nOnly said:v= vt*e^((t*c*2*vt)/m) -vt) / (e^((t*c*2*vt)/m) +1)
where x = (t*c*2*vt)/mv must go to 0
One of the neat things about mathematics is that you usually don't have to guess at the answer. When you get the right answer, you can often tell that you got it right.Ashley1nOnly said:Right because they were all constants, I kept looking at that but the answer is usually always infinity or zero so I was hesitant to pick that.
They want you to find ## F_{engine} ## which will be such that acceleration ## a=\frac{F_{net}}{m} = \frac{F_{engine}-F_{quad}}{m}=0 ## after a long time, and ## v \rightarrow v_T=200 ## km/hour. This means ## F_{engine}=F_{quad} ## after a long time, i.e. once ## v ## approaches its final speed. (##F_{quad} ## actually points opposite ## F_{engine} ## but we included the minus sign when we subtracted them from each other to give ## F_{net} ## ). We know ## F_{quad}=cv^2 =c v_T^2 ## at its final speed, so that ## cv_T^2=F_{engine} ##. ## \\ ## The only problem is that I don't think they gave you the value of the constant ## c ## here. If they did, I didn't see it. ## \\ ## ## c ## should be given as so many Newtons per (km/hour)^2 . That way the answer for ## F_{engine} ## is in Newtons.Ashley1nOnly said:To find the F, I made another post with is a continuation of this if you could help
SammyS said:You have already solved for v. Integrating your expression for acceleration once should just get you back to an expression for v that's equivalent what you already have.
If you'll look back at post 4, you almost had ## v ## in terms of ## t ##, before @kuruman suggested a different method of solution to the differential equation. (You also got ## t ## in terms of ## v ## in post 53 by @kuruman 's method, but solving for ## v ## in terms of ## t ## would take a couple of extra steps. ).Ashley1nOnly said:So I integrate the acceleration above once. a=(cvt^2+cv^2)/m
Why can't I solve for the position, set x0= 0 and x=900 and then solve for t