Example 1:
In this particular example only the x coordinate is used.
The first observation is: at t1=0 object A is at position x1 = 0.
The second observation is: at t1= 10 object A is at position x2 = 20.
Using the equation vx = (x2 - x1)/(t2-t1), shows that vx = (20 - 0)/(10 - 0) = 20/ 10 = 2
This means that object A has a speed 2 in the x direction.
Example 2:
In this particular case also only the x coordinate is used.
The first observation is: at t1=0 object B is at position x1 = 100.
The second observation is: at t1= 10 object B is at position x2 = 80.
Using the equation vx = (x2 - x1)/(t2-t1), shows that vx = (80 - 100)/(10 - 0) = -20/ 10 = -2
This means that object B has a speed -2 in the x direction.
Example 3:
This is a combination of example 1 and example 2.
At t = 0 object A is at position x1 = 0 with a speed vx = 2 and object B is at position x1 = 100 with a speed vx = -2.
That means the two objects approach each other; there will be a collision at position xc
For object A xc = x1 + vx*t = 0 + 2*t.
For object B xc = x1 + vx*t = 100 - 2*t
This gives xc = 0 + 2*t = 100 - 2*t or 4*t = 100 or t = 25
That means there is collision at t = 25 and the distance xc = 50
What is the importance of this whole exercise?
In this whole exercise there are two moving objects involved and based on observations the speed of both objects are calculated. Only one coordinate system is used. And what is also important no direct addition of velocities is used.
This is important because addition of velocities is tricky.
From the point of view of an observer in object A, object B is approaching at a speed of vx = 4. From the point of view of observer B the same is happening: object A is approaching at a speed of vx = 4. This is mathematical correct, but physical wrong. In neither experiment a physical speed of v = 4 is involved. Only a speed of vx =2 (towards the right)
in the x direction and a speed of vx=-2 (towards the left) in the x direction That is the most important physical lesson in the exercise.
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