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# kinematics in 2d help please watch

1. Hey all

I know only hints are allowed, and Id prefer that to be honest, so could I have some hints please?

question 1-

an aeroplane is flying with a velocity vector -90/20 (that is supposed to be vector notation, so please ignore the / which makes it look like a fraction) ms^1.
Find the speed and direction of travel (as a three figure bearing)
***********
Right, when I draw this as a triangle it comes out wierd,

-90 is backwards along a horizontal plane, and the 20 is upwards on a vertical plane right? Drawing that out makes the triangle wierd and longer than higher (If only I could draw the triangles).

Second question:

A boat heads north and 20kph
A wind is blowing SE with a constant speed of 2ms^1

a) Find the magnitude of the speed of the boat
b) Find the bearing
**************

Again, Im having problems with the triangle, drawing it out makes the 2 ms^1 the hypotenuse when, to find the speed shouldnt be the hypotenuse i.e the hypotenuse has to be found?

2. For question 1.

OK, it's a "weird" looking triangle, but it sounds like you have it correct. What's the problem you're having with it?
3. The only calculation to do is work out the missing side (bottom) but I dont see how that gets the speed its flying at as its equals 87 ms^1?
4. (Original post by fishkeeper)
The only calculation to do is work out the missing side (bottom) but I dont see how that gets the speed its flying at as its equals 87 ms^1?
I don't follow you. My interpretation of what you've said is that the bottom is 90, the left side 20, and you need to work out the hypotenuse to find the speed, which will be greater than 90.
5. But that suggests the plane is travelling SW?

Then the bearing would be 270?

I think I may be making this much more confusing than it actually is!
6. (Original post by fishkeeper)
But that suggests the plane is travelling SW?
Why? Your triangle has one side on the left going "vertically up",

Then the bearing would be 270?
A bearing of 270 is due West; it's velocity would be of the form -10/0 say (for example) using your notation.

I think I may be making this much more confusing than it actually is!
I think so too. But I'm not you, and can only go by what you've written. Your initial description of the triangle seemed accurate. You had the two sides, and just needed to work out the length of the hypotenuse to get the speed.

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Updated: February 7, 2010
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