Adding vectors
When adding vector quantities remember that the directions have to be taken into account.
The result of adding vectors together is called the resultant.
In problems, vectors may be added together by scale diagram or mathematically.
Question
A woman walks \(40m\) east then \(30m\) south. The walk takes a total time of \(100s\). Find the total distance travelled.
Because distance is a scalar quantity, the magnitudes of the quantity can just be added together.
Total distance travelled = \(40 + 30 = 70m\)
Question
Now find the resultant displacement.
To do this we need to use a scale drawing. The scale drawing must include a statement of your chosen scale. Where compass points or A direction measured in degrees clockwise from North. Due North is 000, due East is 090, due South is 180 and due West is 270. are included in the question, draw an arrow beside your diagram to indicate north.
Use a large A diagram drawn where each length represents an exact magnitude in direct proportion. Eg, a scale of 1:10 in a diagram means that 1 cm will represent a force of 10 N. for accuracy.
Scale \(1cm\) = \(5m\)
(Mark starting point A, turning point B and finish point C)
Length of vectors consistent with scale
- Draw in first vector (\(40m\) east – on your drawing this will be \(8cm\) long).
- Draw second vector at the end of the first vector (\(30m\) south – this will be \(6cm\) long).
- Now take a straight line from where you started to the end of the second vector – this is your resultant vector.
- Measure the length of the resultant vector – it should be \(10cm\) long. This corresponds to \(50m\) according to your scale. (Line AC in the diagram below).
- Measure the angle BAC (θ below). This is the direction of the resultant displacement. If using compass bearings, remember to relate the angle to 000. For example, in the diagram below, the angle θ is 37˚, so the bearing will be 37 + 90 = 127˚.
When measured on the scale diagram the resultant displacement between points A and C = 50 \(m\) (10 \(cm\) on the diagram)
\(direction\,of\,resultant\,displacement = \theta = 37^\circ S\,of\,E\)
Or mathematically
\(size\,of\,resultant\,displacement = \sqrt {{{(40)}^2} + {{(30)}^2}} = 50m\)
\(\tan \theta = 30 \div 40 = 0.75\)
Direction of resultant displacement
\(\theta = 37^\circ \,S\,of\,E\,or\,a\,bearing\,of\,127\)
Question
Calculate her Average speed = distance / time, over a known distance and time. For example, the average speed of the bus between Edinburgh and Glasgow was 30 m s-1. for this walk.
\(average\,speed = \frac{{distance}}{{time}}\)
\(average\,speed = \frac{{70}}{{100}}\)
\(average\,speed = 0.7\,m\,s^{-1}\)
Question
Now calculate her average velocity for the same walk.
\(average\,velocity = \frac{{displacement}}{{time}}\)
\(average\,velocity = \frac{{50}}{{100}}\)
\(average\,velocity=0\cdot5\,m\,s^{-1}\,on\,a\,bearing\,of\,127\)
You should notice two things:
- Average speed and average velocity can have different magnitudes.
- The displacement and average velocity vectors point in the same direction.
Any vector quantities can be added together like this. It doesn't matter if it is force or velocity, the vectors can still be added together the same way.
Question
A swan is flying due west with a velocity of \(20\,m\,s^{-1}\) and then meets a crosswind of \(8\,m\,s^{-1}\) due north.
By scale diagram, or otherwise, determine the resultant velocity of the swan.
\(Length\, of\, resultant = 10\cdot8 cm\), so velocity size \(= 10\cdot8 \times 2 = 21\cdot 6\, m\, s^{-1}\)
Angle of resultant \(= 22^{\circ}\), so three figure bearing \(= 22 + 270 = 292\).
So the resultant velocity of the swan is \(21\cdot6\,m\,s^{-1}\), at a bearing of \(292\) (or \(22^{\circ}\) north of west)
Or by mathematical calculation:
Size of resultant velocity \(= \sqrt {20^{2} + 8^{2}} = 21\cdot6\, m\, s^{-1}\)
\(\tan \theta = \frac {8}{20} = 0\cdot4,\, \theta = 22^{\circ}\)
So the resultant velocity of the swan is \(21\cdot6\, m\, s^{-1}\), at a bearing of \(292\) (or \(22^{\circ}\) north of west)
