Can you help solve this Speed, Distance, Time Problem?

[Matt97]

Homework Statement

Aircraft A is flying East and maintaining a groundspeed of 340 kt (a kt = speed of 1 NM / hr). Aircraft B is flying in the same direction as aircraft A but 210 NM ahead, maintaining a ground speed of 280 kt. Aircraft A will catch Aircraft B at Point ‘X’. What distance will Aircraft B have travelled when this event occurs?

Relevant Equations

S = D/T etc.

Hi, I'm struggling to solve this problem. Can anyone please solve it and explain it to me?
Thanks.

 

[ZapperZ]

We can't solve it for you. It is against the rules.

We can help you to solve it. But first of all, you need to clarify if you understand the problem, i.e. are you able to read the question and figure out what exactly they're asking for. This is not clear from your short post. So start with explaining a bit more of what you did get from the question, and what you don't get. This is still in the "understanding the question" phase, not the "start to solve" phase.

Zz.

 

[Matt97]

I need to calculate what distance the Aircraft A needs to travel to catch Aircraft B.
What I think I can get from the question is the relationship between the Distance traveled by the two Aircraft.
(340Kts x Time) - 210 = 280Kts x Time
But I'm not sure if that is the correct way to go by it.

Matt

 

[berkeman]

Welcome to the PF.

But I'm not sure if that is the correct way to go by it.

One trick you could use to help you visualize the problem better is to make a 2-D plot of the two aircrafts' position versus time. Draw a 2-D axis with time on the horizontal and distance on the vertical. Put the back (faster) aircraft at the origin, and put the front (slower) aircraft up the vertical axis at it's initial separation from the back aircraft. Then draw the two lines that represent their position versus time (what should the slope of each line be?). You will notice that the bottom/back aircraft has a steeper slope line than the top/front one, and that the two lines cross at some point.

Does that help to get you started?

 

[ZapperZ]

I need to calculate what distance the Aircraft A needs to travel to catch Aircraft B.
What I think I can get from the question is the relationship between the Distance traveled by the two Aircraft.
(340Kts x Time) - 210 = 280Kts x Time
But I'm not sure if that is the correct way to go by it.

Matt

Can you start by at least drawing a picture of the situation?

I don't know if you have been introduce to the concept of "motion diagram" in your lesson. If you have, draw a motion diagram for each of the vessel.

Zz.

 

[Matt97]

Thank you, it does help in visualizing it.
It would take the faster aircraft (340kts) approximately 37 minutes to travel 210Nm then I'm not sure where to go from there.

 

[berkeman]

I didn't open your Word doc (security can be a problem with that file format). If you write it to a PDF file (with a free PDF writer like PrimoPDF), then I'd be happy to take a look.

It would take the faster aircraft (340kts) approximately 37 minutes to travel 210Nm then I'm not sure where to go from there.

What distance will Aircraft B have traveled when this event occurs?
Homework Equations: S = D/T etc.

You wrote an Equation in your first post that relates time and distance and velocity...

 

[Matt97]

Is this equation not relevant to the problem?

 

[berkeman]

Is this equation not relevant to the problem?

What equation? I do see the figure now:

BTW, another way to figure this out is to use the difference in the two speeds, and the initial separation to give you the time to intercept. When you have the time to intercept, that should then give you how far each aircraft has traveled in that time (from their speeds)...

 

[haruspex]

What I think I can get from the question is the relationship between the Distance traveled by the two Aircraft.
(340Kts x Time) - 210 = 280Kts x Time
But I'm not sure if that is the correct way to go by it.

That's the right way to go.
Just rearrange it into the form Time=...

 

[Matt97]

I think I have an answer...
The difference in distance of starting points (210Nm) divided by the difference between the two speeds (60Kts) gives me a time of 3.5 hours.
Multiplying both aircraft speeds (340Kts and 280Kts) by 3.5 gives me 1190Nm and 980Nm respectively.
Subtracting 210Nm from 1190Nm gives me 980Nm so i think the distance traveled by Aircraft A when it passes Aircraft B is 980Nm.

However, if this is correct, I came by it by fluke and I'm having difficulty figuring out why it is the answer.

 

[jbriggs444]

I think I have an answer...
The difference in distance of starting points (210Nm) divided by the difference between the two speeds (60Kts) gives me a time of 3.5 hours.
Multiplying both aircraft speeds (340Kts and 280Kts) by 3.5 gives me 1190Nm and 980Nm respectively.
Subtracting 210Nm from 1190Nm gives me 980Nm so i think the distance traveled by Aircraft A when it passes Aircraft B is 980Nm.

No need to subtract to obtain the 980 nautical miles. You already had it based on the aircraft speed and elapsed time.

However, if this is correct, I came by it by fluke and I'm having difficulty figuring out why it is the answer.

Are you having trouble justifying that 60 knots is the rate at which the faster plane catches up?
Are you having trouble justifying that 3.5 hours is the time to intercept at that rate?
Are you having trouble justifying that 3.5 hours at 280 knots makes 980 nautical miles?

A good way of sanity checking your numbers is to run the calculations based on the value that you obtained.

You decided that the lead aircraft had gone 980 nautical miles at the time of intercept.

Add 210 nautical miles to deduce its final position relative to the trailing aircraft's initial position.

Divide the 980 nautical miles by the leading aircraft's 280 knot speed to obtain the elapsed time.

Multiply that elapsed time by the speed of the trailing aircraft to obtain the distance moved by the trailing aircraft.

Check to see that the final positions of the two aircraft match.

 

[Matt97]

I'm having trouble justifying that 3.5 hours is the tie to intercept at that rate.

 

[haruspex]

I'm having trouble justifying that 3.5 hours is the tie to intercept at that rate.

Then work out where each will be after that time.