RELATIVE MOVEMENT: ONE-DIMENSIONAL, TWO-DIMENSIONAL, EXERCISES - PHYSICAL - 2025

The relative motion of a particle or an object is that observed with respect to a particular reference point that the observer has chosen, which can be fixed or in motion. Velocity always refers to some coordinate system used to describe it.

For example, the passenger of a car in motion and who travels comfortably asleep in his seat is at rest relative to the driver, but not for an observer standing on the sidewalk who sees the car go by.

Figure 1. Airplanes maintain a certain speed relative to each other when practicing stunts. Source: Pixabay.

Then the movement is always relative, but it happens that in general the coordinate or reference system is chosen having its origin in the Earth or the ground, a place considered stationary. In this way the concern is focused on describing the movement of the object under study.

Is it possible to describe the speed of the sleeping copilot compared to a passenger traveling in another car? The answer is yes. There is freedom to choose the value of (x _o, y _o, z _o): the origin of the reference system. The selection is arbitrary and depends on the preference of the observer, as well as the ease it provides for solving the problem.

Relative motion in one dimension

When the movement takes place along a straight line, the mobiles have speeds in the same direction or in the opposite direction, both seen by an observer standing on Earth (T). Does the observer move relative to the mobiles? Yes, with the same speed that they carry, but in the opposite direction.

How does one mobile move with respect to the other? To find out, the velocities are added vectorially.

-Resolved example 1

With reference to the figure shown, indicate the relative speed of car 1 with respect to car 2 in each situation.

Figure 2. Two cars are going on a straight road: a) in the same direction and b) in opposite directions.

Solution

We will assign a positive sign to the speeds to the right, and a negative sign to the left. If a mobile goes to the right at 80 km / h, a passenger on this mobile sees the observer on Earth move at - 80 km / h.

Suppose everything happens along the x-axis. In the following figure the red car is moving at +100 km / h (seen from T) and is about to pass the blue car traveling at +80 km / h (also seen from T). How fast does a passenger in the blue car approach the red car?

The labels are: v _1/2 speed of car 1 with respect to 2, v _{1 / T} speed of car with respect to T, v _{T / 2} speed of T with respect to 2. Vector addition:

v _1/2 = v _{1 / T} + v _{T / 2} = (+100 km / h - 80 km / h) x = 20 km / h x

We can do without the vector notation. Notice the subscripts: multiplying the two on the right you should get the one on the left.

And when they go the other way? Now v _{1 / T} = + 80 km / h and v _{2 / T} = -100 km / h, therefore v _{T / 2} = + 100 km / h. The passenger of the blue car will see the red car approach:

v _{1/2 =} v _{1 / T} + v _{T / 2} = +80 km / h +100 km / h = 180 km / h

Relative motion in two and three dimensions

In the following diagram, r is the position of the plane seen from the xyz system, r 'is the position from the x'y'z' system and R is the position of the system with a prime with respect to the system without prime. The three vectors form a triangle in which R + r '= r, therefore r ' = r - R.

Figure 3.- The plane moves with respect to two coordinate systems, in turn one of the systems moves with respect to the other.

Since the derivative with respect to time of the position is precisely the velocity, it results:

v '= v - u

In this equation v 'is the speed of the plane with respect to the x'y'z' system, v is the speed with respect to the xyz system and u is the constant speed of the prime system with respect to the unprimed system.

-Solved exercise 2

An airplane is going north with airspeed of 240 km / h. Suddenly the wind begins to blow from west to east at a speed of 120 km / depending on the earth.

Find: a) The speed of the plane with respect to the ground, b) The deviation experienced by the pilot c) The correction that the pilot must make to be able to aim directly north and the new speed with respect to the ground, once the correction has been made.

Solution

a) There are the following elements: plane (A), ground (T) and wind (V).

In the coordinate system in which north is the + y direction and the west-east direction is + x, we have the given speeds and their respective label (subscripts):

v _{A / V} = 240 km / h (+ y); v _{V / T} = 120 km / h (+ x); v _{A / T} =?

The proper vector sum is:

v _{A / T} = v _{A / V} + v _{V / T} = 240 km / h (+ y) + 120 km / h (+ x)

The magnitude of this vector is: v _{A / T} = (240 ² + 120 ²) ^1/2 km / h = 268.3 km / h

b) θ = arctg (v _{A / V} / v _{V / T}) = arctg (240/120) = 63.4º North of East or 26.6º Northeast.

c) To continue north with this wind, you must point the bow of the plane to the northwest, so that the wind is pushing it directly to the north. In this case the speed of the plane seen from the ground will be in the + y direction, while the speed of the plane with respect to the wind will be northwest (it does not necessarily have to be 26.6º).

By Pythagorean theorem:

α = arctg (v _{V / T} / v _{A / T}) = arctg (120 / 207.8) = 30º Northwest

-Solved exercise 3

It takes a person 2 minutes to walk down a stationary escalator. If the ladder works, it takes the person 1 minute to go down while standing still. How long does it take for the person to walk down with the ladder running?

Solution

There are three elements to consider: the person (P), the ladder (E) and the ground (S), whose relative speeds are:

v _{P / E}: speed of the person with respect to the ladder; v _{I / O}: speed of the ladder with respect to the ground; v _{P / S}: speed of the person with respect to the ground.

As seen from the ground by a fixed observer, the person descending the ladder (E) has a velocity v _{P / S} given by:

v _{P / S} = v _{P / E} + v _{I / S}

The positive direction is going down the ladder. Let t be the time it takes to walk down and L the distance. The magnitude of the person's velocity v _{P / S} is:

v _{P / S} = L / t

t ₁ is the time it takes to walk down with the ladder stopped: v _{P / E} = L / t ₁

And t ₂ the one it takes to go down still on the moving stair: v _{E / S} = L / t ₂

Combining the expressions:

L / t = L / t ₁ + L / t ₂

Substituting numerical values ​​and solving for t:

1 / t = 1 / t ₁ + 1 / t ₂ = 1/2 + 1/1 = 1.5

So t = 1 /1.5 minutes = 40 seconds.

References

1. Bauer, W. 2011. Physics for Engineering and Sciences. Volume 1. Mc Graw Hill. 84-88.
2. Figueroa, D. Physics Series for Sciences and Engineering. Volume 3rd. Edition. Kinematics. 199-232.
3. Giancoli, D. 2006. Physics: Principles with Applications. 6 ^th. Ed. Prentice Hall. 62-64.
4. Relative Motion. Recovered from: courses.lumenlearning.com
5. Wilson, J. 2011. Physics 10. Pearson Education. 166-168.