Hypothetical vehicle speed

[Iustitia]

I have a problem that is completely hypothetical and revolves around the speed of a non-existent aerial vehicle. Using the following specifications, what speed can this hypothetical flying vehicle move?

Standard weight: 43.4 metric tons
Max Acceleration: 0.93 G
Rocket Thrusters: 2 x 24000 kg , 4 x 1870 kg

I know that speed = distance / time. I have the weight of the object in question, and I believe I can derive its mass from its weight to get 43,400 kilograms. I'm not very good with mathematical concepts in physics but from what I understand 'G' is a unit of acceleration and 1 standard unit of G or gravity is 9.8 m/s2. So a max acceleration of 0.93 G would be 9.114 m/s2.

If force = mass * acceleration then 9.114 m/s2 * 43,400 kg = 395547.6 Newtons of force. And that's great, but, I'm not asked for the force, but the speed. I don't know where to go from here and the information listed above is all I have. The equations I've done don't seem to help any. Any aid or direction I can get here would be appreciated. Thank you.

 

[phinds]

Well, if you maintain a constant acceleration of .93G, then it will go faster and faster right up to the point where it falls apart.

 

[jh0]

But of course, the acceleration of the vehicle cannot be 0.93 G all the time. As soon as the vehicle starts moving, assuming it does not move in a vacuum, the aerodynamic drag will oppose the thrust from the rockets. The maximum speed will be reached with the balance in all the forces acting on the vehicle: thrust, gravity and drag (maybe I forget something).

 

[phinds]

But of course, the acceleration of the vehicle cannot be 0.93 G all the time. As soon as the vehicle starts moving, assuming it does not move in a vacuum, the aerodynamic drag will oppose the thrust from the rockets. The maximum speed will be reached with the balance in all the forces acting on the vehicle: thrust, gravity and drag (maybe I forget something).

While I agree w/ what you are saying in practice, it is still true that if you had enough power and DID maintain a constant acceleration, it WOULD keep going faster and faster (against drag, as you point out) until it falls apart (BECAUSE of the forces set up by the drag).

My point was to alert him to the fact that the problem as he stated it is not well formulated. For example, there is zero information in the statement that would let you figure what the drag IS, so on that basis alone the problem has no solution.

 

[jack action]

You are taking the problem the wrong way. The acceleration is unknown at first (you cannot spit out a number like that).

First you find the force you have:

2 * 24000 + 4 * 1870 = 55480 kgf

A force of 55480 kgf is equivalent to a force of 544259 N (see any online unit converter to go from kilogram-force to Newton).

THEN you find your acceleration:

544259 / 43400 = 12.54 m/s² (= 1.28 G)

Now, say you are initially at rest (v = 0 m/s) and you are keeping this acceleration for 1 s:

0 + 12.54 * (1) = 12.54 m/s

So after 1 s your vehicle will have reached 12.54 m/s ( = 45 km/h).

Like it was mentioned by others, some other forces should react (usually the drag force is the predominant one and it varies with speed). You have to evaluate that force at that new speed. Say that you find out that it is equal to 4000 N. Then the new force is:

544259 - 4000 = 540259 N

The new acceleration is:

540259 / 43400 = 12.45 m/s²

And you new speed after another 1 s of travel will be:

12.54 m/s + 12.45 m/s² * (1 s) = 24.99 m/s

And you repeat the process until your acceleration = 0. At that point, you have reach you terminal velocity.

To get more precision, you can reduce the 1 s interval such that the speed increase is smaller, but you will need more calculations. Usually you need small intervals at the beginning, but as the acceleration decreases, you can increase the time interval (even greater than 1 s).

And if you are in space and that you have no opposing force, then you will have a constant acceleration of 12.54 m/s² as long as you have fuel. Replace the 1 s interval by the time it takes to empty your tank and you will get your terminal velocity.