# Thread: The point of no return

1. ## The point of no return

Thanks for your attention, appreciate it.

2. Originally Posted by hungrybarts

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 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.
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.

3. 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

4. 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.
..

5. 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.

6. finally i figured this out here is my working

i suppose it has covered 1-3

now the last problem is 4