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# Cie Additional Math- velocity vector questions

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1. I'm got an additional math exam coming up in a few weeks and although revision on most of the content is going fairly well I'm still having a lot of difficulty with the vectors section - specifically the velocity vector questions with bearings and the crossing the river type questions. Here's one question from an exam paper along my own workings . Any help will be much appreciated...
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Just replying for now, so you know it's not being ignored.

+rep for presenting your work so clearly.

I think if we can get the diagram correct, then that will go a long way to resolving your issues.

I'm stuck for time this morning, so can't pick this up properly until lunch time, assuming no one else has resolved things by then.
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So, plane is flying directly from A to B, to its velocity relative to the ground will be along the line A to B. (diagram 1)

Now, velocity of plane relative to ground is the vector sum of velocity of the plane relative to air plus velocity of air relative to the ground.

Note also that the wind is blowig from the East, so it's going West.

See diagram 2 for the velocity triangle, with the bearing the plane steers marked as "a"

Can you take it from there?

4. (Original post by ghostwalker)
So, plane is flying directly from A to B, to its velocity relative to the ground will be along the line A to B. (diagram 1)

Now, velocity of plane relative to ground is the vector sum of velocity of the plane relative to air plus velocity of air relative to the ground.

Note also that the wind is blowig from the East, so it's going West.

See diagram 2 for the velocity triangle, with the bearing the plane steers marked as "a"

Can you take it from there?

Thank you so much . Yes , using trig I managed to get the answer just fine. I always have a problem with placing the directions properly even with the sketching. Also I can't understand that if the plane is going from A to B then why isn't the plane's velocity at where you placed the resultant vector and vice versa since its the plane that going from A to B .

*Oh sorry I just thought about the question and what I said a bit more and it states that the bearing of B from A is 030° but not the direction that the pilot had to steer the plane in , which of course is the first question we're asked to solve. I really need to read these questions more carefully.

I'm starting to think that perhaps its not that I can't understand the concept but that I have a problem applying it right. Which I guess is part of not understanding the concept correctly .

Anyways thank you so much for your help. I'll have a go at a few more questions. I see that you decided to use the trig approach to solving this question but is it possible for you to solve it using unit vectors? The unit vectors approach just seems a bit less complicated to me , but hopefully now that I know I need to read the questions more carefully then maybe I can just stick to using the trig approach. Thanks again .
I see that you decided to use the trig approach to solving this question but is it possible for you to solve it using unit vectors? The unit vectors approach just seems a bit less complicated to me , but hopefully now that I know I need to read the questions more carefully then maybe I can just stick to using the trig approach. Thanks again .
The use of unit vectors here would make things more complicated, IMO.

Something like:

Velocity of wind = -80i

Velocity of plane relative to air = 320 sin a i + 320 cos a j

So, vel. plane relative to ground = (320 sin a -80)i + (320cos a ) j

This vector needs to be on a bearing of 030 degrees, so:

sin 30 =(320 sin a -80) / (320cos a )

and solve for a.

I'd stick with a good diagram and trig.
6. I think I'm beginning to understand the concept better since I got further in this question but I got stumped right near the end . I hope you can clarify where I went wrong .
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I think I'm beginning to understand the concept better since I got further in this question but I got stumped right near the end . I hope you can clarify where I went wrong .
Perhaps if you could explain what your diagram represents, what each line is for.
Check it against the question.
8. (Original post by ghostwalker)
Perhaps if you could explain what your diagram represents, what each line is for.
Check it against the question.
Well if the ocean liner is traveling on a bearing of 090° then I figured that means its heading east, and I indicated that in my diagram. I think you're trying to point out that I've gotten the direction wrong again. When I've changed the diagram to the following,
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Well if the ocean liner is traveling on a bearing of 090° then I figured that means its heading east, and I indicated that in my diagram. I think you're trying to point out that I've gotten the direction wrong again. When I've changed the diagram to the following,
Note thtat the liner is on a bearing of 315 from the lifeboat (it looks like you have it the other way around, but can't be sure). And the liner is moving east.

If you plot their positions at 0600, you should be able to work out the position of the liner at 07:30.

The lifeboat will rendevous with the liner at that position, and the lifeboat is heading directly to that position.

Time to eat - then going out.
10. (Original post by ghostwalker)
Note thtat the liner is on a bearing of 315 from the lifeboat (it looks like you have it the other way around, but can't be sure). And the liner is moving east.

If you plot their positions at 0600, you should be able to work out the position of the liner at 07:30.

The lifeboat will rendevous with the liner at that position, and the lifeboat is heading directly to that position.

Time to eat - then going out.
Yes , I think I really do have it the other way around (sigh) , its part of me being directionally challenged I reckon. Well Bon Appetite and thanks again.

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