# Trajectory in Physics: Characteristics, Types, Examples and Exercises

The trajectory in physics is the curve that describes a piece of furniture as it passes through successive points during its movement. As this can adopt numerous variants, the paths that the cell phone can follow.

To get from one place to another, a person can follow different paths and in different ways: walking along sidewalks on streets and avenues, or arriving by car or motorcycle on a highway. During a walk through the forest, the hiker can follow a complicated path that includes turns, ascending and descending levels, and even passing the same point several times.

If the points the phone is traveling through follow a straight line, the path will be straight. This is the simplest path because it is one-dimensional. Specifying the position requires a single coordinate.

But the cell phone can follow a curvilinear trajectory, being able to be closed or opened. In these cases, tracking position requires two or three coordinates. These are movements in plane and space respectively. This has to do with links : material conditions that limit movement. Some examples are:

– The orbits that describe the planets around the sun are closed paths in the shape of an ellipse. Although, in some cases, they can approximate a circular, as in the case of Earth.

– The ball that the goalkeeper kicks in a goal kick follows a parabolic path.

– A bird in flight describes curved trajectories in space, because, in addition to moving in an airplane, it can go up or down at will.

The trajectory in physics can be expressed mathematically when the moving position is known at any time. Let R be the position vector, which in turn has the coordinates X , e and Z in the general case of a three-dimensional motion. By knowing the function r (t), the trajectory will be completely determined.

## Types

Generally speaking, the trajectory can be a pretty tricky curve, especially if you want to express it mathematically. Therefore, it starts with the simplest models, where cell phones travel in a straight line or in an airplane, which can be the one on the floor or any other suitable one:

### Movements in one, two and three dimensions

The most studied trajectories are:

– Rectilinear when traveling straight in a horizontal, vertical or inclined. A ball thrown vertically upwards follows this path or an object that slides downhill along an incline. They are one-dimensional movements, just one coordinate is enough to completely determine their position.

– Parabolic , in which the cell phone describes a parabola arc. Frequently, any object thrown obliquely under gravity (a projectile) follows this path. To specify the movable position, two coordinates must be given: x and y.

– Circular , occurs when the moving particle follows a circle. It is also common in nature and in daily practice. Many everyday objects follow a circular path, such as tires, machine parts and orbiting satellites, to name a few.

Related:   Bohr’s Atomic Model: Characteristics and Postulates

– Elliptical , the object moves after an ellipse. As stated at the beginning, it is the path that the orbiting planets follow around the sun.

– Hyperbolic astronomical objects under the action of a central force (gravity) can follow elliptical (closed) or hyperbolic (open) paths, which are less frequent than the first.

– Helical or spiral movement , like that of a bird rising in a thermal current.

– Swing or pendulum , the cell phone describes an arc in back and forth movements.

## Examples

The trajectories described in the previous section are very useful for getting a quick idea of ​​how an object moves. Anyway, it is necessary to clarify that the trajectory of a cell phone depends on the location of the observer. This means that the same event can be seen in different ways, depending on where each person is.

For example, a girl is pedaling at a constant speed and throws a ball. She notes that the ball follows a straight trajectory.

However, to an observer standing on the road and watching it pass, the ball will have a parabolic motion. For him, the ball was initially launched with an inclined speed, the result of acceleration by the girl’s hand plus the speed of the bicycle.

### Explicit, implicit and parametrically moving path

– Explicit , directly specifying the curve or geometric location given by the equation y(x)

– Implicit , where a curve is expressed as f (x, y, z) = 0

– Parametric , in this way the x, y and z coordinates are given based on a parameter that, in general, is chosen as the time t . In this case, the trajectory is composed of the functions: x (t), y (t ) and z (t).

Next, two highly studied kinematics trajectories are detailed: the parabolic trajectory and the circular trajectory.

#### Inclined launch in vacuum

An object (the projectile) is thrown at an angle a with horizontal velocity and initial velocity or as shown in the figure. Air resistance is not taken into account. The movement can be treated as two independent and simultaneous movements: one horizontal with constant velocity and the other vertical under the action of gravity.

x (t) = x o + v ox .t

y (t) = y o + v oy .t -½g.t 2

These equations are the parametric equations of projectile launches. As explained above, they have the parameter t , which is time, common.

The following triangle can be seen in the right triangle:

ox = v or cos θ i

oy = v o sin θ i

Replacing those equations that contain the launch angle into the parametric equations results in:

x (t) = x or + v or cos θ i .t

y (t) = y o + v o . sin θ i .t -½g.t 2

##### Parabolic Path Equation

The explicit trajectory equation is found by clearing t from the equation by x (t) and substituting it into the equation for y (t). To facilitate the algebraic work, it can be assumed that the origin (0,0) is located at the launch point and therefore x o = y o = 0. This is the equation of the path explicitly .

#### circular path

A circular path is given by:

(x – x o ) 2 + (y – y o ) 2 = R 2

Here x or y the represent the center of the circle described by the mobile and R is the radius of it. P(x,y) is a point on the path. In the shaded right triangle (figure 3), note that:

x = R. cos θ

y = R. sin θ

The parameter in this case is the sweep angle θ, called the angular displacement. In the specific case where the angular velocity ω (scan angle per unit of time) is constant, it can be stated that:

θ = θ or + ω t

Where θ o is the initial angular position of the particle, which is considered to be 0, it is reduced to:

θ = ω t

In this case, time returns to parametric equations, such as:

x = R.cos ω t

y = R. sin ω t

The unit vectors i and j are very convenient for writing the position function of an object r (t). They indicate directions on the x- axis and the y- axis , respectively. In your terms, the position of a particle that describes a uniform circular motion is:

r (t) = R.cos ω i + R. sin ω j

## solved exercises

### Exercise solved 1

A cannon can fire a bullet with a speed of 200 m / s and an angle of 40° from the horizontal. If launching on flat ground and air resistance is neglected, find:

a) The equation of the trajectory y (x) ..

b) The parametric equations x (t) and y (t).

c) The horizontal range and duration of the projectile in the air.

d) The height at which the projectile is when x = 12,000 m

#### Solution a)

a) To find the trajectory, the values ​​given in the y(x) equation from the previous section are substituted: y (x) = tg 40°. x – {9.8 / (2 ‘400 2. cos 2 40) } x 2 ⇒ y (x) = 0.8391 x – 0.0000522x two

#### Solution b)

b) The launch point at the origin of the coordinate system (0,0) is chosen:

x (t) = x or + v ox. t = 400 ´ cos 40º.t = 306.42. t.

y (t) = y o + v oy .t -½g.t 2 = 400 ′ sin 40º.t – 0.5 ´ 9.8 ´ 2 = 257.12 t – 4.9.t 2

#### Solution c)

c) To find the time the projectile lasts in the air, e (t) = 0 , with the launch being made on flat ground:

0 = 257.12.t – 4.9.t 2

t = 257.12 / 4.9 s = 52,473 s

The maximum horizontal range is replacing this value in x (t):

max = 306.42 ′ 52.47 m = 16077.7 m

Another way to find x max directly is by setting y = 0 in the trajectory equation:

0 = 0.8391 x max – 0.0000522 x max

x = 0.8391 / 0.0000522 m = 16078.5 m

There is a small difference due to rounding of decimal places.

#### Solution d)

d) To know the height when x = 12000 m, this value is directly substituted in the trajectory equation:

y (12000) = 0.8391 ‘12,000 to 0.0000522 ‘12000 2 m = 2552.4 m

### Solved exercise 2

The position function of an object is given by:

r (t) = 3t i + (4-5t 2 ) j m

To locate:

a) The trajectory equation. What curve is this?

b) The initial position and the position when t = 2 s.

c) The displacement performed after t = 2 s.

#### Solution

a) The position function was given in terms of the unit of vectors i and j , which determine the direction in respectively the x and e axes , therefore:

x (t) = 3t

y(t) = 4 -5t 2

The path equation y(x) is found by clearing t from x(t) and replacing y(t):

t = x / 3

y(x) = 4-5. (X / 3) 2 = 4 – 5 x 2 /9 (parabola)

b) The initial position is: r (2) = 4 j m ; the position at t = 2 s is r (2) = 6 i -16 j m

c) The displacement r is the subtraction of the two position vectors:

Δ r = r (2) – r (2) = {6 i -16 j }  4 j = 6 i – 20 j m

### Solved exercise 3

The Earth has a radius R = 6300 km and it is known that the period of rotation of its movement around its axis is one day. To locate:

a) The equation of the trajectory of a point on the Earth’s surface and its position function.

b) The speed and acceleration of that point.

#### Solution a)

a) The position function for any point on the circular orbit is:

r (t) = R.cos ω i + R.sen ω j

You have the Earth’s radius R, but not the angular velocity ω, but you can calculate the period, knowing that, for circular motion, it’s valid to say that:

ω = 2π × frequency = 2π / period

The movement period is: 1 day = 24 hours = 1440 minutes = 86400 seconds, therefore:

ω = 2π / 86400 s = 0.000023148 ​​​​s -1

Replacing in position function:

r (t) = R.cos ω + R. sin ω j = 6300 (cos 0.000023148t i + sin 0.000023148t j ) Km

The parametric trajectory is:

x (t) = 6300. cos 0.000023148t

and (t) = 6300. sin 0.000023148t

#### Solution b)

b) For circular motion, the magnitude of the linear velocity v of a point is related to the angular velocity w by:

v = ω R = 0.000023148 ​​​​s -1 ´ 6300 Km = 0.1458 Km / s = 145.8 m / s

Even though it is a movement with a constant velocity of 145.8 m / s , there is an acceleration that points to the center of the circular orbit, responsible for keeping the point in rotation. It is the centripetal acceleration for c , given by:

c = v 2 / R = (145.8 m / s) 2 / is 6300 × October 3 m = 0.00337 m / s 2 .