RELATIVE MOTION

When we examine the movement of a body or object, we always describe its movement in relation to a particular frame or reference system. For example, let's say we are going to say goodbye to a friend who is about to travel by bus. As the bus moves away from the stop, we observe that our friend moves away from the bus. However, for him things are perceived differently, the bus is still and it is us who are moving away from the bus stop.

So, are we the ones moving or is he our friend? Incredible as it may seem, we both move, only the difference in our observations arises from our different frames of reference.

For our purpose, a reference system is a physical object to which we attach our coordinate system. In everyday life, that object is the ground. An example is when a policeman measure the speed of drivers on a road or motorway, they always make the measurement in relation to the ground. If the officer were moving while making the measurement, he would get a very different value for the speed.

Relative motion in one dimension

Let's continue with the previous example, the policeman (at the origin of reference system A) is parked at the side of a road, watching car P pass quickly. Maria (at the origin of reference system B) is driving along the same road at a constant speed and is also looking at car P.

Relative motion in one dimension
Figure 1. Relative movement in one direction

Let us suppose that both measure the position of the car in a time t, therefore, it is fulfilled that:

\ large x_ {PA} = x_ {PB} + x_ {BA} \ hspace {2.0em} (1)

This equation tells us that the position x PA of the particle P measured by the observer A (the police), is equal to the position x PB of P measured by B (Maria), plus the position of B measured by A. Notice in the equation that the first subscript represents what is observed and the second is who is observing.

If we derive equation (1) with respect to time, we obtain:

\ large \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left (x_ {PA} \ right) = \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left ( x_ {PB} \ right) + \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left (x_ {BA} \ right)

Therefore, the horizontal components of velocity are related by:

\ large v_ {PA} = v_ {PB} + v_ {BA} \ hspace {2.0em} (2)

This equation does not express that speed vPA of P, measured by A, is equal to the speed vPB of P measured by B plus velocity VBA of B measured by A. This last term is the velocity of frame B relative to frame A.

Here we only consider frames of reference that move at constant speed relative to each other. In our example, this means that Maria (reference system B) moves at a constant speed vBA relating to the police (reference system A). The car (our P particle) can change its speed and direction, that is, accelerate.

To find the relationship of the acceleration of the car measured by Maria and the police, we derived equation (2) with respect to time.

\ large \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left (v_ {PA} \ right) = \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left ( v_ {PB} \ right) + \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left (v_ {BA} \ right)

As VBA is constant, we get:

\ large a_ {PA} = a_ {PB}

Which means in other words that the acceleration of a particle measured by an observer in a reference system is the same as that measured by any other moving with constant speed in relation to the first system.

Relative movement in two directions

Suppose that our observers are again at the origin of the reference systems A and B, while B moves at a constant speed vBA relative to A, looking at the motion of a particle P. In addition, suppose that the axes of both frames or systems remain in parallel throughout the observation. Figure 2 shows the movement of P at time t.

Relative motion in two dimensions
Figure 2.

At that time, the origins of the reference frames are separated by a distance: \vec{r}_{BA} The position vector of the particle P in relation to the origin of system A is:

\ vec {r} _ {PA}

and with respect to the origin of B is:

\ vec {r} _ {PB}

From the figure we can see that the three vectors are related to each other through the expression:

\ large \ vec {r} _ {PA} = \ vec {r} _ {PB} + \ vec {r} _ {BA} \ hspace {2.0em} (3)

By deriving equation (3) with respect to time, we can relate the velocities y of the relative particle P to our observers:

\ large \ vec {v} _ {PA} = \ vec {v} _ {PB} + \ vec {v} _ {BA} \ hspace {2.0em} (4)

Equations (3) and (4) are known as Galilean transformation equations and relate position and velocity as measured by observers in relative motion.

Now if we derive equation (4) with respect to time, we can relate the accelerations of A and B, and of the particle P relative to our observers. However, we must bear in mind that it is constant, therefore:

\ large \ vec {a} _ {PA} = \ vec {a} _ {PB}

Therefore, just like in one dimension, we have to:

When two observers in different frames or reference frames move at constant speed relative to each other, they will measure the same acceleration for a moving P particle.

Relative motion exercises

  1. In Figure 1, suppose that María moves at a constant speed relative to the police vAB= 40km / h and the carriage P moves in the negative direction to the x-axis.
  • If the police measure a constant speed vAB= -85km / h for car P, what is the speed that Maria measures?

We know that:

\ large v_ {PA} = v_ {PB} + v_ {BA}

Thus:

\ large \ begin {align *} v_ {PB} & = v_ {PA} -v_ {BA} \\ & = -85 \ tfrac {km} {h} -40 \ tfrac {km} {h} \\ & = -125 \ tfrac {km} {h} \ end {align *}

  • If the car P brakes relative to the police at t = 10s at a constant acceleration, what is its acceleration aPA regarding the police?

Since the acceleration is constant, we know that:

\ large v = v_ {0} + at

We know that the initial speed relative to the police is vAB= -85km / h and the final speed is 0, so we have:

\ large \ begin {align *} a & = \ frac {v-v_ {0}} {t} = \\ & = \ frac {0- \ left (-85 \ tfrac {km} {h} \ right) } {10s} \ cdot \ frac {1000m} {1km} \ cdot \ frac {1h} {3600s} = \\ & = \ frac {85 \ tfrac {km} {h}} {10s} \ cdot \ frac { 1 \ tfrac {m} {s}} {3.6 \ tfrac {km} {s}} = \\ & = 2.4 \ tfrac {m} {s ^ {^ {2}}} \\ \ end {align *}

  • What is the acceleration toPB of car P relating to Maria during braking?

We know that the initial velocity of P relative to Maria is:

\ large v_ {PB} = 125 \ tfrac {km} {h}

The final speed of P relative to Maria is -40km / s, because it is the speed relative to her movement when the car stops. Thus:

\ large \ begin {align *} a & = \ frac {v-v_ {0}} {t} = \\ & = \ frac {-40 \ tfrac {km} {h} - \ left (-125 \ tfrac {km} {h} \ right)} {10s} \ cdot \ frac {1000m} {1km} \ cdot \ frac {1h} {3600s} = \\ & = \ frac {85 \ tfrac {km} {h} } {10s} \ cdot \ frac {1 \ tfrac {m} {s}} {3.6 \ tfrac {km} {s}} = \\ & = 2.4 \ tfrac {m} {s ^ {^ {2}}} \\ \ end {align *}

That confirms what was said previously, because the police and María move at constant relative speed, both measure the same acceleration for the car.

  1. A car travels east with a speed of 60km / h. Raindrops fall with a constant speed vertically from Earth. The traces of the rain on the side windows of the car make an angle 30 ° with the vertical. Find the speed of the rain in relation to the car and the Earth.
Relative movement exercise
Figure 3.

The relationship between speeds is:

\ large \ vec {v} _ {gT} = \ vec {v} _ {AT} + \ vec {v} _ {gT} \ hspace {2.0em} (5)

Where:

  • VgT is the speed of the drop relative to the Earth.
  • VAT speed of the car relative to the Earth.
  • VgA is the speed of the drop relative to the car.

We need to decompose the vectors into their components in the coordinate system of Figure 3 and then solve equation 5.

For the y coordinates, we have:

\ large \ begin * align *} {v} _ {gT_ {y}} & = {v} _ {AT_ {y}} + {v} _ {gA_ {y}} \\ {v} _ {gT} & = 0 + {v} _ {gA} \ cos \ theta \\ {v} _ {gT} & = {v} _ {gA} \ cos \ theta \ hspace {2.0em} (6) \ end {align *}

Now, for the x coordinates, we have:

\ large \ begin * align *} {v} _ {gT_ {x}} & = {v} _ {AT_ {x}} + {v} _ {gA_ {x}} \\ 0 & = {v} _ { AT_ {x}} + {v} _ {gA} \ sin \ theta \\ {v} _ {gA} & = - \ frac {{v} _ {AT_ {x}}} {\ sin \ theta} = \\ {v} _ {gA} & = - \ frac {80 \ tfrac {km} {h}} {\ sin 30 ^ {\ circ}} = \\ & = - 160 \ tfrac {km} {h} \ end {align *}

Thus:

\ large \ vec {v} _ {gA} = 160 \ tfrac {km} {h} \ hspace {0.3em} to \ hspace {0.3em} 30 ^ {\ circ} \ hspace {0.3em} west

Now, substituting this value in equation (6), we have:

\ large v_ {gT} = \ left (-160 \ tfrac {km} {h} \ right) \ cos 30 ^ {\ circ} = - 138.6 \ tfrac {km} {h}

That is, the speed of the relative drop with the Earth is:

\ large \ vec {v} _ {gT} = 138.6 \ tfrac {km} {h} \ hspace {0.3em} towards \ hspace {0.3em} al \ hspace {0.3em} south