The point of no return

http://img375.imageshack.us/img375/1...oreturnqv0.jpg

I have absolutely no idea about this. Can you give me a step by step solution?
Thanks for your attention, appreciate it.

Quote:

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Originally Posted by hungrybarts

http://img375.imageshack.us/img375/1...oreturnqv0.jpg

I have absolutely no idea about this. Can you give me a step by step solution? Mr F says: You want to be given a step-by-step solution to an assessment task (Speechless) Do the words academic fraud mean anything to you? I will give you some help with question 1 (see main post) - then you're on your own.
Thanks for your attention, appreciate it.

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Let d be distance from the air field and t be time flying.

On an appropriate set of axes, draw the line d = 350 t (where has the 350 come from?) but stop at a point. Label that point (d1, t1). t1 is the time flown with the wind. What is the relationship between d1 and t1 (you will need this relationship)?

Now draw a line from the point (d1, t1) to the point (0, 4) (where has the 4 come from?) What must the slope of this line be? Use the value of this slope to help calculate a value of t1.

Hence answer question 2.

my apologize for that mr fantastic but apparently, no one in my class is able to solve this so, should the distance be on the y-axis and the time be on the x-axis (distance against time)? and i still don't get how to do no 1

Quote:

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Originally Posted by hungrybarts

my apologize for that mr fantastic but apparently, no one in my class is able to solve this so, should the distance be on the y-axis and the time be on the x-axis (distance against time)? Mr F says: Yes.

and i still don't get how to do no 1 Mr F says: Now try doing what I suggested in my first post.

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Hello,
I believe I can answer the question. However, as I understand a "hint" is given to help someone, but its usage is not imperative. I have so decided to disregard the graph hint and thought up another solution, whose correctness I cannot guarantee.

My idea is the following:
We are looking for a maximum distance which we can travel given the circumstances and come back.
So let S1 be the distance we are travelling forward, V1 - the going-forward speed, t1 - the time we are travelling forward and S2, V2, t2 - the same variables only that when we are travelling back.

So our aim is to make S1 = S2.
The formula from the physics classes we know for a distance travelled is:
S = V / t.
So S1 = V1 / t1 and S2 = V2 / t2.
We need S1 to be equal to S2, so
V1 / t1 = V2 / t2.
We know that t1+t2 = 4 and V1 and V2's values are known.
From here onwards it is all simplification and eventually I get that the maximum time that can be travelled forward is 7/3 hours and 5/3 hours back.

As I said I cannot guarantee that the solution is correct and I am open to any critics as long as they do not leave the borders of logic and clear thinking.

finally i figured this out here is my working

http://img227.imageshack.us/img227/3...thgraphrb7.jpg

i suppose it has covered 1-3

now the last problem is 4