## Relative Motion

## Galilean Relativity

Consider two observers, A and B, moving relative to each other. How can we compare their observations?

In particular, consider two reference frames, labeled A and B, which move relative to each other with constant velocity v, oriented so they are parallel and their x-axes align with their relative velocity. The velocity of frame B relative to A is v and the velocity of frame A relative to B is &endash; v. Further, to simplify matters, we start our clocks or stopwatches when the origins of the two reference frames coincide. All this is shown schematically below.

Now consider an object located at point

P. Observers in both frames record its position and velocity. Of course, they get different numbers but these numbers can be related to each other by using these transformation equations:x _{A}= x_{B}+ vty

_{A}= y_{B}v

_{Ax}= v_{Bx}+ vv

_{Ay}= v_{By}where x

_{A}and y_{A}are the coordinates of pointPmeasured by observer A while x_{B}and y_{B}are the coordinates measured by observer B; v_{Ax}, and v_{Ay}are the components ofP's velocity measured by A, and v_{Bx}and v_{By}are the velocity components measured by B. Everybody measures t for the time and v is the relative speed of the two observers. These equations are the Galilean Relativity Transformations. The position equations may be combined and written in vector form asr_{A}=r_{B}+vtwhile the velocity equations may be combined and written in vector form as

v_{A}=v_{B}+vAll these transformation equations may be seen in the following schematic diagram:

We expect Newton's

FirstLaw, theLaw of Inertia, to be valid forbothobservers. We expect both these frames to beinertialframes. If Newton'sLaw of Inertiais valid,allthe other laws of Mechanics--and ideas like conservation of energy, conservation of momentum, and conservation of angular momentum--will also be valid.The relative motion need not be along one of the axes, of course. Consider a point P that is located by displacement vector

r_{p1}with respect to reference frame 1 and located by displacement vectorr_{p2}with respect to reference frame 2. And reference frame 2 is located by vectorr_{21}. This is shown here, in this sketch,These displacement vectors are related by

Notice the details of the subscripts,

We can then take the time derivative of this equation to find the speed of point P as seen by the two reference frames,

This equation corresponds to the diagram and assumes that we know the velocity

v_{p2}in reference frame 2 and the relative velocityv_{21}. If, instead, we know the velocityv_{p1}in reference frame 1 and the relative velocityv_{21}, we can simply solve forv_{p2},If the two reference frames move with

constant relative velocity, then taking the time derivative one more time providesThat is, the acceleration seen by observers in the two reference frames is the same.

ExampleA boat heads due North across a river with a speed of v

_{br}= 10 km/h relative to the water. The river has a speed of v_{rE}= 5 km/h due East, relative to Earth. Determine the velocity of the boat relative to Earth v_{bE}(that is, the velocity of the boat relative to an observer watching from the bank of the river).We can write the appropriate relative velocity equation,

v_{bE}=v_{br}+v_{rE}v

_{bE}= SQRT [ v_{br}^{2}+ v_{rE}^{2}]v

_{bE}= SQRT [ ( 10 )^{2}+ ( 5 )^{2}] km / hv

_{bE}= 11.2 km / h= tan

^{-1}(^{opp}/_{adj}) = tan^{-1}( 5 / 10 ) = tan^{-1}(0.50) = 26.6^{o}

ExampleNow we want to cross the same river with the same boat. But, this time, we want to go directly across the river. At what angle, upstream, must we head so that our velocity relative to Earth,

v_{bE}, is directly across the river. What is our new speed v_{bE}, relative to Earth (or our land lubber observer sitting on the bank)? Our vector equation is stillv_{bE}=v_{br}+v_{rE}but our vector

diagramlooks quite different now,The

magnitude(or lenght) of vectorv_{br}is still 10 km/hr.v _{br}^{2}= v_{bE}^{2}+ v_{rE}^{2}v

_{rE}^{2}= v_{br}^{2}- v_{bE}^{2}v

_{rE}^{2}= [ ( 10 )^{2}- ( 5 )^{2}] km/hrv

_{rE}= SQRT [ ( 10 )^{2}- ( 5 )^{2}] km/hrv

_{rE}= 8.66 km / hr= tan

^{-1}(^{opp}/_{adj}) = tan^{-1}( 5 / 8.660 ) = tan^{-1}( 0.577 ) = 30^{o}

Return, for a moment, to our two observers A and B.Observers

AandB"see" different things. But both agree that theLaws of Physicsare the same.What do they observe for the

speed of light?

Experimentally,both observers find the same value for the speed of light!How can this be? Stay tuned . . . until we reach Modern Physics and Special Relativity. Many interesting and unexpected things wait for us along the journey.

AccelerationRelative Motion at High SpeedsReturn to ToC, Ch4, Two-Dimensional Motion(c) Doug Davis, 2001; all rights reserved