How far will a toy car travel with a small rocket motor

[jack action]

Now what do you mean by "I can find vf at t=1s by trial and error"?

I meant you try a value of $v_f$ and check if it gives 1 for $t$. If not, you try another value until you find $v_f$ that makes $t=1$. I suggested that method because, at first glance, I thought I couldn't isolate $v_f$ as a function of $t$. But you could. This function backward:
$$t = -\frac{1}{2CB}\ln\left|\frac{v_f-C}{v_f+C}\right|$$
is this:
$$v_f = C\frac{1 + ±e^{-2CBt}}{1 - ±e^{-2CBt}}$$
And I don't know if it will always be the case (probably), but for the example I did, the $±$ is actually $-$, so the the correct signs are:
$$v_f = C\frac{1 - e^{-2CBt}}{1 + e^{-2CBt}}$$

 

[marciokoko]

Yeah thanks, I had actually started mounting it on excel. I have the following issue though, according to my numbers I get for e^(-2*B*C*t) a very small number x10^-11, so 1- or 1+ that simply equals 1 and thus vf = C. Here are my values:

F = 12.8N
m = 0.3kg
Crr= 0.0385
g=9.8m/s2
p=1.23 kg cm3
Af=16cm2
t = 1s
Cd = 1.28

I get:
A=42.289
B=3.779
C=3.345

Do you get the same values?

 

[jack action]

$B$ should be $\frac{1.23 \times 1.28 \times 0.0016}{2 \times 0.3} = 0.0041984$ and $C = 100.363$.

 

[marciokoko]

Ok, I had read kg/cm3 for p. Ok so I have 0.004 for B and 105 for C.That gives me 40.17 for vf. So that is the terminal velocity of the car, meaning the speed its going after the burn?

That gives me 20.6 for distance traveled during the burn and 376 after for a total of 396m. That seems like a lot, 1,298 ft. Id like to put it in context, taking it back to the vertical rocket scenario. The rocket is lighter, I think they weigh about 170 g and the datasheet says they rise to about 750 ft. According to my calculations the car would travel 2x the distance (almost). Anyway, this means I shouldn't do this on my street since its not 396m long, I don't think. And someone mentioned straight-path travel which of course has been a concern because I know any slightest misalignment of the rocket and itll go sideways, end up hitting something and go nowhere near the 396m.

 

[jack action]

Id like to put it in context, taking it back to the vertical rocket scenario. The rocket is lighter, I think they weigh about 170 g and the datasheet says they rise to about 750 ft. According to my calculations the car would travel 2x the distance (almost).

The rocket doesn't have rolling resistance. Instead, it has gravity to fight back. That is the same thing as having $C_{rr} = 1$; That is 26 times greater than your rolling resistance. Putting this number and a weight of 170 g into the equations, I get a total distance of 91 m or 4 times less (I had 370 m instead of your 396 m). And I suspect the mass of the rocket decreases noticeably as the fuel burns, which would give a greater acceleration and increase the total distance.

The point is, it is normal that it travels (a lot) further against rolling resistance than against gravity. When the rocket stops burning, the deceleration is 1g due to gravity; It is only 0.0385g due to rolling resistance (that's what $C_{rr}g$ means in the equations).

 

[marciokoko]

Thank you so much for everything. I've got the physics part cleared up. Now since I'm curious I'll take a peak at the calculus again. I love it!

I know it's a other question but first I want to understand the concept of how when you define the problem as dv = a * dt. You want to find how v changed with time, which is of course by definition, a. That is a derivative, right?

And then you solve it by integrating it. And I remember from my calculus class that integrals and derivatives are opposites. I don't think I ever made the connection between the integral being the solution to a derivative. I just remember doing tons of: take the derivative of x2...2x, OK next...

Conceptually it's very nice because derivatives are the model for the change in one vrlariable with respect to another, v and dt. And integrals are the infinitesimal sum of those differences? Or something like that?

 

[jack action]

Yes, you got the basic concept.

If an integral becomes too complex to solve for you, you can also do hundreds of calculations on an Excel sheet by using small changes $\Delta t$ instead of infinitesimal changes $dt$. If $\Delta t$ is small enough, then the following is true:
$$\frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t} \approx \frac{dv}{dt} = a$$
So if you start knowing that $v_1$ = $t_1$ = 0, you define a small $\Delta t$, say 0.1 s. Knowing $v_1$, you can calculate the acceleration with the equation I've given you earlier, such that:
$$v_2 = v_1 + a \Delta t$$
Now you know the velocity at time $t_2$. You can recalculate the acceleration with the new found velocity and find $v_3$ based on $v_2$ and the new acceleration. And so on and so forth, until you reach an acceleration of 0 or the end of the burning time or whatever condition you have.

To get the distance travel, it is the same thing. For example, for point 2:
$$x_2 = x_1 + v_{avg}\Delta t$$
Where the average velocity $v_{avg}$ during $\Delta t$ is:
$$v_{avg} = \frac{v_1 + v_2}{2}$$
The smaller will be your $\Delta t$, the more precise should be your answer.

 

[sophiecentaur]

Without knowing the Force / time characteristic of the engine, I don't understand how you can calculate the distance traveled during burn. You can. at least, calculate how far it will go after the burn if you believe the Impulse value (and if you can characterise the track and the drag etc).

 

[Nugatory]

Without knowing the Force / time characteristic of the engine, I don't understand how you can calculate the distance traveled during burn. You can. at least, calculate how far it will go after the burn if you believe the Impulse value (and if you can characterise the track and the drag etc).

The thrust-time curves for these Estes motors are (not surprisingly, now that I think about it) available on the internet. I got to say... The internet does make some things a lot easier than when I was a kid.

 

[sophiecentaur]

The thrust-time curves for these Estes motors are (not surprisingly, now that I think about it) available on the internet. I got to say... The internet does make some things a lot easier than when I was a kid.

I never used one but I imagine, without the figures, the only information you could have was a whoosh that peaked near the end? But, there again, as a kid, just sending it way up in the air was all the reward you needed.