math query-do not send me to any web site plz?

Question

]am way out of school.]
a train weighs 120,000 pounds
it can travel, using diesel fuel, 16 kilometers
in one hour

if it switches power, it can travel 4 kilometers
in one hour using elect generators

if it again switches power, to wind power,
it can travel 2 kilometers per hour.

if the engineer wishes to incorporate
all 3 means of propulsion, what speed or
distance will he be able to reach?

offer answer and foruma.

Answer 1

I don't remember the formulas to resolve this the proper way, but if you can ignore some conversion loose, and despite the fact that the wind power will vary as the train goes faster (if the wind generator is on the train).
I have to say that the energy of the three systems together will be the sum of all the energy generated, having more energy you can conver it to more force or more speed depending on what you need.
In other words, as the amount of joules of energy will be the sum of all of them, I assume that the speed will also be and the answer should be:

16km/h + 4km/h + 2km/h = 22km/h

Answer 2

You don't state the size of the diesel tank, nor the battery capacity. The problem states rates, only. If it were to say, "...diesel... 16 km FOR one hour", etc. then it might make more sense.

I'll assume these are rates for the given hour (then fuel runs out or batteries run out).

It is implied that he cannot use all 3 means simultaneously. For diesel and battery power (16+ 4): 20 km / hour for the first 2 hours. That's an average speed of 10 km/ hr.

After that he's at the mercy of the wind. So long as the wind keeps up, he can keep the train moving along slowly, at 2 km /hour indefinitely. After the 3rd hour his average speed will be 22 km/3 hours = 7.3333 km/ hour. After 4 hours it will be 24 km / 4 hours = 6 km/hr. Then 26 km / 5 hours... I think you get the idea.

.

Answer 3

The information is not in the question. You state 4km per hour on elect. power only. But will the elec motor run faster than 16km with no load? the elec might only run 4km at its fastest speed even if it is not driving the locomotive. if you turn it on the elec with the diesel it would slow down.
The speeds are not additive
In a perfect world and with each power source capable of infinite unloaded speed (or much higher than 16km/hr) you could calculate the amount of power that each system produces in use. The power is additive and if all powers were summed them divided by the load you could calculate the speed produced by the total power.
But the speed of the wind is probably not much higher (if any) than the diesel speed. So it might not add anything.

Answer 4

If it really wants to go far it should use its best fuel as long as it can. The train's weight, which I think is improbably low since a diesel locomotive weighs 200,000 pounds by itself, is not really a factor in the answer.

First it should use its diesel fuel to travel at 16 kph until it is out of fuel. Since we don't know how much fuel it has on board we can't give any more info on distance.

Next it should connect to the overhead electric lines and go until it runs out of areas with electric lines. That would be at 4 kph and we have no idea what distance.

Then since it is out of fuel and electric it can travel using wind at 2 kph. How long the wind will blow no one knows so that is another mystery.

All in all you are short some key data for getting an answer.

Answer 5

A very complex question, and most of it refers to impractical ideas. You don't provide enough information to come anywhere close to a real answer.

For example, how would it use "elect[ric?] generators" without consuming diesel fuel or other fuel in an engine to power the generators? (That is the way most diesel locomotives work, of course.)

So there really is no way to understand or answer your question without lots and lots more specifics.

But at least I didn't send you to a web site. Aren't you glad?

Answer 6

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Answer 7

It'll go forever. You'll have to put some kind of limitation on the engineer before you can come up with any other answer.

Answer 8

23 mph km=d/t*1.5