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.

Printable View

- August 9th 2008, 04:05 PMhungrybartsThe 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. - August 9th 2008, 07:20 PMmr fantastic
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. - August 10th 2008, 11:47 PMhungrybarts
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

- August 11th 2008, 02:54 AMmr fantastic
- August 11th 2008, 03:25 AMLogic
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. - August 13th 2008, 12:28 AMhungrybarts
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