M1 Vector Help: Find Walker's Position Using i and j Vectors

[CathyLou]

Hi.

I'm really stuck on the following M1 level vector question so any help would be really appreciated.

Unit vectors i and j are directed due east and north respectively.

A mountain rescue post O receives a distress call via a mobile phone from a walker who has broken a leg and cannot move. The walker says that he is by a pipeline and he can also see a radio mast which he belives to be south-west of him. The pipeline is known to run north-south for a long distance through the point with position vector 6i km, relative to O. The radio mast is known to be at the point with position vector 2j km, relative to O.

(a) Using the information supplied by the walker, write down his position vector in the form (ai + bj).

Thank you.

Cathy

 

[chaoseverlasting]

Any ideas on how to go about it? What does a vector describe? How does that help you?

 

[learningphysics]

I believe the question wants you to assume that the mast is 45 degrees west of south from the walker...

So draw a picture... use a regular cartesian coordinate system. O is the origin. Draw the pipe as a line... and draw a point where the mast is...

Do you see how to use the 45 degree angle?

 

[CathyLou]

Thanks for your help.

Sorry, but I'm still not sure how to start the calculations? Could you please help me through the stages?

Thank you.

Cathy

 

[learningphysics]

Did you draw a picture? What is the x coordinate of the walker?

 

[CathyLou]

Did you draw a picture? What is the x coordinate of the walker?

Yeah, I drew a picture. Is the x coordinate 6?

Cathy

 

[learningphysics]

Yeah, I drew a picture. Is the x coordinate 6?

Cathy

Yes, exactly. The man says he can see the mast south west... that means that the man is north east of the mast... So draw a line from the mast going north east, at an angle of 45 degrees above the horizontal... the place where this line intersects the pipeline is where the man is located...

You already know the x coordinate... you just need to find the y coordinate... you'll have to use a little geometry here...

 

[CathyLou]

Awesome, thanks for your help!

I got (6i + 8j).

Cathy

 

[learningphysics]

Awesome, thanks for your help!

I got (6i + 8j).

Cathy

Looks good.

 

[CathyLou]

Hi.

I'm stuck on this next part too. Could you please give me some tips?

The rescue party moves at a horizontal speed of 5 km/h. The leader of the party wants to give the walker an idea of how long it will take for the rescue party to arrive.

Calculate how long it will take for the rescue party to reach the walker's estimated position.

Thank you.

Cathy

 

[learningphysics]

Hmmm... I'm wondering what exactly they mean by horizontal speed.

Assuming they move in a straight line from the origin to the man... just take the distance and divide by speed...

You can get the distance from the man's position vector which you calculated in the last part (distance is just the magnitude of the vector)... just divide that by the given speed.

 

[CathyLou]

Thank you for your help.

I got that the time is 2 hours.

Could someone please help me with the last part (part d)? I would really appreciate it.

The rescue party sets out and walks straight towards the walker's estimated position at a constant horizontal speed of 5 km/h. After the party has traveled for one hour, the walker rings again. He is very apologetic and says that he now realizes that the radio mast is in fact north-west of his position.

(c) Find the position vector of the walker.

I got this as (6i - 4j).

(d) Find, in degrees to one decimal place, the bearing on which the rescue party should now travel in order to reach the walker directly./B]

Thank you.

Cathy

 

[learningphysics]

Find the position of the rescue party after the 1 hour... Then what you need to do is calculate the angle of the vector joining the rescue party and the walker (who's at 6i-4j)

So if you draw north-south and east-west axes through the point that is the rescue party's position... you can calculate the angle of the vector from one of these 4 axes (whichever you choose)...

depending on which angle you choose to calculate, your answer will be given differently... for example if you calculate the angle from the east axis... then your answer would be x degrees north of east... or x degrees south of east... or if you calculate the angle from the south axis, your answer would be x degrees east of south, or x degrees west of south... etc...

I think all these are equally valid... maybe your text or professor has a preference for how to give the bearing...

 

[CathyLou]

Thank you for all your help!

I got a final answer of 139.4 degrees from the north.

Cathy

 

[learningphysics]

Thank you for all your help!

I got a final answer of 139.4 degrees from the north.

Cathy

Hmmm... I'm getting 159.4 degrees east of north...