# Relative Velocity; Help needed please

"Position along the N/S axis" corresponds to what most might call the "y-coordinate" and it is the same for both ships:
View attachment 78918

think i'm on the right track, will let you know if I get it solved, thanks for all your help so far!

"Position along the N/S axis" corresponds to what most might call the "y-coordinate" and it is the same for both ships:
View attachment 78918
Think I have it, the minimum value of the j (verticle/y component) is 5j (5km/h) and if the value of V=6 then the directions of interception are due north and N6.59degreesW

Nathanael
Homework Helper
Think I have it, the minimum value of the j (verticle/y component) is 5j (5km/h) and if the value of V=6 then the directions of interception are due north and N6.59degreesW
If it is going due north at 6 m/s it will go too far and not crash. You want the Northern component of velocity of B to be 5m/s then it will always eventually crash (with one exception) because they will always have the same N/S coordinate.

If it is going due north at 6 m/s it will go too far and not crash. You want the Northern component of velocity of B to be 5m/s then it will always eventually crash (with one exception) because they will always have the same N/S coordinate.

by intercept in the question I think they mean when ship B crosses the path which ship A is taking, in the case of a collision or going to take in the case of an interception? because 2.(b)(ii) asks that if v=6 show that B can travel in either of two directions to intercept A and find these directions.

The minimum value asked in part (i) is when they will collide but in part (ii) they are referring to the ships crossing paths I believe?

Nathanael
Homework Helper
by intercept in the question I think they mean when ship B crosses the path which ship A is taking, in the case of a collision or going to take in the case of an interception? because 2.(b)(ii) asks that if v=6 show that B can travel in either of two directions to intercept A and find these directions.

The minimum value asked in part (i) is when they will collide but in part (ii) they are referring to the ships crossing paths I believe?
I doubt it. If this is what they mean then there are infinite different directions which ship B could travel. But to crash, there are exactly 2 angles which it can travel.

I doubt it. If this is what they mean then there are infinite different directions which ship B could travel. But to crash, there are exactly 2 angles which it can travel.

Yes, i see, well how do I go about calculating the direction is v=6

Nathanael
Homework Helper
Yes, i see, well how do I go about calculating the direction is v=6
For which directions is the component of the velocity in the northern direction equal to 5 m/s?

For which directions is the component of the velocity in the northern direction equal to 5 m/s?

is E33.56N one of them?

Nathanael
Homework Helper
is E33.56N one of them?
Yep. And the other?

Yep. And the other?
I want to say 90-33.56 but I have a feeling that is wrong, other than that I really don't know

Nathanael
Homework Helper
E33.56N
By E33.56N you mean "33.56 degrees East of North," right?

Is it possible for it to travel west of north and crash? How many degrees west of north would it need to go?

By E33.56N you mean "33.56 degrees East of North," right?

Is it possible for it to travel west of north and crash? How many degrees west of north would it need to go?

Yeah thats what I meant, I should have expressed it as N33.56E, I get where you're coming from but I still don't know how to go about calculating it

Nathanael
Homework Helper
Yeah thats what I meant, I should have expressed it as N33.56E, I get where you're coming from but I still don't know how to go about calculating it
Well, I don't know how else to explain it except to just tell you. It would be the same angle as you calculated, 33.56°, except west of north. Draw (or imagine) a picture to understand it better. The angle is measured against the y axis. No matter if you measure it to the left of the y axis or to the right of the y axis, it still gives you the same y component (5 m/s)

Well, I don't know how else to explain it except to just tell you. It would be the same angle as you calculated, 33.56°, except west of north. Draw (or imagine) a picture to understand it better. The angle is measured against the y axis. No matter if you measure it to the left of the y axis or to the right of the y axis, it still gives you the same y component (5 m/s)

I thought about that but then I thought it was too simple, thanks so much for all your help! It seems so simple now that I know, makes me feel stupid for not seeing it earlier lol, thanks again!