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now what to do if the 2 vessels are not on the same axis .... ??
again, as is so often the case DRAW the diagrams
- a relative position vector diagram
- a relative velocity diagram.
often 1 diagram will suffice but with the 2 sets of information on the graph.
from these identify as many of the angles and sides as the question or indeed the vectors directions supply.
As always the next thing to do in these questions is to determine the Relative Velocity between the 2 objects
the next thing would be to attempt to solve the question that we were asked,
find the shortest distance between the the bodies in the subsequent motion.
then we can by using position vectors, work out the distance between them, this makes one side of the triangle
draw the Relative velocity vector and by knowing the other angle between the 2 lines is at a right angle
What can make life easier is to move both particles until they are on the same i or j line, i.e. they are left / right or up / down from each other.
If you are going to follow this path make sure that you move both bodies the same amount of time.
Alternatively we can use the coordinate geometry equation for the perpendicular distance between a point and a line,
so by inference of the line of the relative velocity we need to determine the slope and the location of the 1st moving body.
use this as the line ax + by + c = 0 and the point to which all this is relative as x1 y1,
There are other ways, such as moving both on in time so they are on the axis, but be careful that the shortest distance was probably back in time!
This is best if you take one of the cars to the junction, preferable if you take the observing car, move them both on and then calculate the distance between the cars, the angle of the Vab the shortest distance between the relative moving object and the observer is to be found by drawing a line perpendicular to Vab and to the observer.
I personally think you should try all 3 methods .... even though my least favourite method of solving these is the coordinate geometry method, it clearly is the simplest, and most thorogh.
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