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Unit 1
Vectors
Topics Covered:
Any vector is space can be defined by its coordinates in terms of its x, y, and z position.
In a two dimensional frame we need only consider its x and y co-ordinates. Therefore any vector can be thought of as a resultant when its horizontal component in the x direction is added to its vertical component in the y direction.
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Figure 1: |
Figure2: |
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Figure 3: |
Figure 4: |
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Figure 5: |
We can also define Vx and Vy in
terms of V and
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The vector equation for vector v
in terms of |
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Here are some examples of vector components:
1. What are the horizontal and vertical components of a plane moving at 300 km/h [E650N]?
Refer to Figure 5 above.
Here q = 65 0 and v = 300 km/h
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2. What are the components of a vector 8 cm long with bearings [W150S]?
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Addition of Vectors using the components method
When adding two or more vectors using the components method follow these guidelines.
| The animations below illustrate the fact that irrespective of the path one takes from an initial position to a final position the sum of all vectors will be equal to a resultant vector which has only two components: an x (horizontal) component and a y (vertical component). Here the vectors a, b, and c all have their own x and y components which can be added together to give a total horizontal component V1 and a total vertical component V2. Thes, in turn can be added to find out the resultant vector VTot. Note that in both scenarios (figure A and figure B), the end point with respect to the origin is the same. | |
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Figure A |
Figure B |
Example:
Find the final displacement of a man who first walks North for 5.0 km,
then East for 3.0 km and finally [E600S] for 8.0 km. The sketch
is not to scale.
Solution:
Given: d1 = 5.0 km [N], d2 = 3.0 km [E], d3 = 8.0 km [E600S]
Find:
All components
dtx
dty
dt
q, the direction of the final displacement vector dt
Method:
set-up a table for each vector where you can enter the x and y components of
each.
| Vector | Sketch | X component (horizontal) | Y component (vertical) |
| d1 = 5.0 km [N] |
d1x = 0 |
d1y = +5 |
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| d2 = 3.0 km [E] |
d2x = + 3 |
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d3 = 8.0 km [E600S] |
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Totals |
dTx = 0 + 3 + 4 |
dTy = 5 + 0+ (-)6.9 |
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| Use Pythagoras' Theorem
to find the magnitude of total displacement :
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| DIRECTION |
q = inv tan (dTy/dTx) = 150 |
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Conclusion:
The total displacement of the man is : 7.3 km [E150]