Integration help on physics problem

[Charles Link]

ln| vt/ (vt-v) |
ln| (1-vt/v)|

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]

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)]

 

[Ashley1nOnly]

t= m/(c*2*vt) *ln [ (vt+v )/ (vt-v)]

 

[Charles Link]

Without checking it carefully, I believe what you have is correct. Now as $t \rightarrow +\infty$, what must $v$ do to have the right side give $+\infty$?

 

[SammyS]

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)]

Please read posts (#46) and (#49), and save yourself some work.

It's fine to practice your algebra, but ...

 

[Ashley1nOnly]

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

 

[Ashley1nOnly]

How do I calculate the F?

 

[SammyS]

How do I calculate the F?

Look at the OP (original post):
line 3 or 4 under

The Attempt at a Solution

,
assuming F is what you initially had as F_C .

 

[Charles Link]

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

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]

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.

 

[Charles Link]

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.

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]

To find the F, I made another post with is a continuation of this if you could help

 

[Charles Link]

To find the F, I made another post with is a continuation of this if you could help

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]

to find out how long it takes to get to the destination

a= (cvt^2+cv^2)/m

integrate it twice to get position and then solve for t

 

[Charles Link]

The time it takes to get to the destination will be approximately $t=(900 \, miles)/(200 km/hour)$. $\\$ For the exact time that it takes, you need to set $900 \, miles=\int\limits_{0}^{t} v \, dt$, and solve for $t$. $\\$ And note, they change units on you=it is 900 "miles".

 

[Ashley1nOnly]

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

 

[SammyS]

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.

 

[Ashley1nOnly]

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.

Right so all he did was integrate the velocity again to get position and then solved for t. the initial x is zero and final x is 900 miles

 

[Charles Link]

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

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. ).
Your first way actually gets the formula you need here. Your notation was clumsy in post 4, confusing $F_{engine}$ with $F_c$ and $F_{engine}$ multiplied by $c$, and you also made an algebraic error or two, so let me give you what you should have: $v=v_T \, \tanh(At)$ where $A=\frac{F_{engine}}{m v_T}$. (If you solved post 53 for $v$, you would also get this result). Meanwhile $F_{engine}=c v_T^2$, but they never told you what to use for $c$. $\\$ The $v=v_T \tanh(At)$ function can be integrated once to get the distance of 900 miles. (It's a standard integral that looks something like $\ln|\cosh(x)|$). You could solve for $t$ as I said in post 63. $\\$ Note that the acceleration $a=\frac{dv}{dt }$ is not constant, so $v^2=v_o^2+2ad$ does not work. $\\$ And yes, you could integrate the acceleration twice. We just integrated it once to get this formula for $v$ in post 4, and we need to integrate it once more, as described above. In a very simple form $\int \tanh(x) \, dx=\int \frac{sinh(x)}{\cosh(x) } \, dx=\int \frac{d \, (cosh(x))}{cosh(x)}=\ln|\cosh(x)|$. The constant $A$ just needs to be included. $\\$ Suggestion: Make the 900 miles into 900 km, and use a value of (Note: edited from c=100 to c=0.2 ) $c=0.2 \ Newtons/(km/hour)^2$. The value of $c$ isn't going to affect the final result significantly for the time $t$. Compute $t$ exactly, as described in post 65, and compare to the estimate at the top of post 65.

 

[Charles Link]

And for this one, let me simply write out what I got and you can try to verify it: Integrating $s=\int\limits_{0}^{t} v \, dt=\int\limits_{0}^{t} v_T \tanh(At) \, dt=\frac{v_T}{A} \ln|\cosh(At)| |_0^t =\frac{v_T}{A} \ln|\cosh(At)|=\frac{mv_T^2}{F_{engine}} \ln|\cosh(\frac{cv_T}{m}t)|=\frac{m}{c} \ln|\cosh(\frac{cv_T}{m}t)|$ . $\\$ Now $\cosh(x)=\frac{e^x+e^{-x}}{2}$. For large $x$ , $\ln|\cosh(x)| \approx x$. $\\$ The result is $s \approx (\frac{m}{c}) (\frac{cv_T}{m} t) =v_T t$, confirming the first line of post 65 as a good estimate. $\\$ And @fresh_42 Probably good that you closed the other thread. In this post 70, I simply presented the result of the integration. Otherwise, we might need a few more posts to get there.