## Wednesday, May 2, 2018

### Unizor - Physics - Mechanics - Kinematics - Frame of Reference - Relativ...

Notes to a video lecture on http://www.unizor.com

Relative Motion

The subject of this topic is motion of one object relative to another. In particular, assume that we have an original inertial frame of reference and two objects moving with constant velocities in this frame.

Consider further that we would like to use another frame of reference -
the one associated with one of the objects and axes parallel to the axes
of the original frame of reference.

As we have proven before, this new frame of reference is inertial. Our task now is to determine the vectors of position and velocity of objects in this new frame of reference.

As a typical example, consider two trains, A and B, going in opposite
directions along two straight lines parallel to each other with constant
speeds.

Our original frame of reference is the ground. We choose an X-axis along
the train lines with positive direction coinciding with the direction
of train A. Then Y- and Z-coordinates of positions and velocities of
both trains are zero.

The task at hand is to determine how fast train B passes train A from the viewpoint of a passenger of train A.

Obviously, translated this task to a language of frames of reference, we
have to determine the characteristics of motion of the train B in an
inertial frame associated with train A.

Let Pa(t) and Pb(t) be position vectors
of our two objects, A and B, in the original frame of reference. These
positions, obviously, depend on time since objects A and B are moving.

Their velocity vectors are, correspondingly, Va and Vb, which we assume are constant.

Vector Pab(t) represents the position vector of object B in the new frame of reference associated with object A.

Since, according to the rules of vectors' addition,

Pb(t) = Pa(t) + Pab(t)

we can find Pab(t):

Pab(t) = Pb(t)Pa(t)

Differentiating this by time, we obtain the corresponding equations for velocity vectors:

Vb = Va + Vab

Vab = VbVa

The equations above, for positions and velocities, determine the motion
of object B in an inertial frame of reference associated with object A
and express the Galileo's Relativity Principles.

Let's apply this rule to a task with two trains, A and B, moving in opposite directions along the X-axis with speeds

Va = 100 km/hour and

Vb = −90 km/hour

(minus for train B signifies that, if the positive direction of the
X-axis is towards the movement of train A, train B moves in an opposite
direction).

Then the relative speed of train B for a passenger on train A is

Vab = VbVa =

= −90 − 100 = −190

The minus sign means that a passenger on train A sees train B moving
backwards (relative to his own movement, which is towards positive
direction of the X-axis). The absolute value of a relative speed in this
case of motion in opposite directions, as we see, is a sum of absolute
values of corresponding speeds.

Let's assume that the trains move in the same (positive) direction of
the X-axis with above speeds. Now their velocity vectors are

Va = 100 km/hour and

Vb = 90 km/hour

Then the relative speed of train B for a passenger on train A is

Vab = VbVa =

= 90 − 100 = −10

The minus sign means that a passenger on train A sees train B moving
backwards (because train A is faster and, therefore, train B, relative
to a passenger on train B, moves backward). The absolute value of a
relative speed in this case of motion in the same direction, as we see,
is a difference of absolute values of corresponding speeds.

Other examples of velocities' arithmetic are as follows.

1. A platform A moves uniformly along straight railing with speed Va = 3 m/sec. A person B moves on the platform in the same direction with speed Vab = 1 m/sec relative to a platform. What is the person's speed Vb relative to the ground?

Let's associate our original frame of reference with the ground with X-axis directed along the direction of a moving platform.

Since

Vb = Va + Vab

Vb = 3 + 1 = 4 (m/sec)

2. A river A flows uniformly along straight banks with speed Va = 3 m/sec. A boat B moves in the water with speed Vab = 10 m/sec. What is the boat's speed Vb relative to the ground when it moves down the river flow and against it?

Let's associate our original frame of reference with the ground with X-axis directed along the flow of the river.

Then Va = 3 m/sec.

If the boat goes down the river, its speed relative to water is Vab = 10 m/sec.

If the boat goes up the river, its speed relative to water is Vab = −10 m/sec.

From the same equation for speeds

Vb = Va + Vab

follows that, when the boat goes down the river, its speed is

Vb = 3 + 10 = 13 (m/sec)

and up the river

Vb = 3 − 10 = −7 (m/sec)

(here minus means that the boat moves against the positive direction of the X-axis, associated with the flow of the river).