DIFFERENCES BETWEEN SPEED AND SPEED (WITH EXAMPLES) - PHYSICAL - 2025

The differences between speed and speed exist, although both are related physical quantities. In common language one term or the other is used interchangeably as if they were synonyms, but in Physics it is necessary to distinguish them.

This article defines both concepts, points out the differences, and explains, using examples, how and when one or the other is applied. To simplify we consider a particle in motion and from there we will review the concepts of speed and speed.

Figure 1. Speed ​​and velocity of a particle moving in a curve. Prepared by: F. Zapata.

Differences between speed and speed

	Speed	Speed
Definition	It is the distance traveled per unit of time	It is the displacement (or change of position) in each unit of time
Notation	v	v
Mathematical object type	Climb	Vector
Formula (for a finite period of time) *	v = Δs / Δt	v = Δr / Δt
Formula (for a given instant of time) **	v = ds / dt = s' (t)	v = dr / dt = r '(t)
Explanation of the formula	* Length of the path traveled divided by the period of time used to travel it. ** In instantaneous speed, the period of time tends to zero. ** The mathematical operation is the derivative of the path arc as a function of time with respect to the instant t of time.	* Vector displacement divided by the time period in which the displacement occurred. ** At instantaneous speed the time lapse tends to zero. ** The mathematical operation is the derivative of the position function with respect to time.
characteristics	To express it, only a positive real number is required, regardless of the spatial dimensions in which the movement occurs. ** Instantaneous speed is the absolute value of instantaneous speed.	It may take more than one real number (positive or negative) to express it, depending on the spatial dimensions in which the movement occurs. ** The modulus of instantaneous velocity is instantaneous speed.

Examples with uniform speed on straight sections

Various aspects of speed and speed were summarized in the table above. And then, to complement, several examples are considered that illustrate the concepts involved and their relationships:

- Example 1

Suppose a red ant moves along a straight line and in the direction indicated in the figure below.

Figure 2. An ant on a straight path. Source: F. Zapata.

In addition, the ant moves uniformly so that it travels a distance of 30 millimeters in a period of time of 0.25 seconds.

Determine the speed and velocity of the ant.

Solution

The speed of the ant is calculated by dividing the distance Δs traveled by the time period Δt.

v = Δs / Δt = (30 mm) / (0.25s) = 120 mm / s = 12 cm / s

The speed of the ant is calculated by dividing the displacement Δ r by the time period in which the displacement was made.

The displacement was 30 mm in the 30º direction with respect to the X axis, or in compact form:

Δ r = (30 mm ¦ 30º)

It can be noted that the displacement consists of a magnitude and a direction, since it is a vector quantity. Alternatively, the displacement can be expressed according to its Cartesian components X and Y, in this way:

Δ r = (30 mm * cos (30º); 30 mm * sin (30º)) = (25.98 mm; 15.00 mm)

The speed of the ant is calculated by dividing the displacement by the time period in which it was made:

v = Δ r / Δt = (25.98 mm / 0.25 s; 15.00 mm / 0.25 s) = (103.92; 60.00) mm / s

This velocity in Cartesian components X and Y and in units of cm / s is:

v = (10.392; 6.000) cm / s.

Alternatively the velocity vector can be expressed in its polar form (modulus ¦ direction) as shown:

v = (12 cm / s ¦ 30º).

Note: in this example, since the speed is constant, the average speed and the instantaneous speed coincide. The modulus of the instantaneous velocity is found to be the instantaneous speed.

Example 2

The same ant in the previous example goes from A to B, then from B to C and finally from C to A, following the triangular path shown in the following figure.

Figure 3. Triangular path of an ant. Source: F. Zapata.

Section AB covers it in 0.2s; the BC runs it in 0.1s and finally CA runs it in 0.3s. Find the mean speed of the trip ABCA and the mean speed of the trip ABCA.

Solution

To calculate the average speed of the ant, we begin by determining the total distance traveled:

Δs = 5 cm + 4 cm + 3 cm = 12 cm.

The time span used for the entire journey is:

Δt = 0.2s + 0.1s + 0.3s = 0.6 s.

So the mean speed of the ant is:

v = Δs / Δt = (12 cm) / (0.6s) = 20 cm / s.

Next, the average speed of the ant in the ABCA route is calculated. In this case, the displacement made by the ant is:

Δ r = (0 cm; 0 cm)

This is because the offset is the difference between the end position minus the start position. Since both positions are the same, then their difference is null, resulting in a null displacement.

This null displacement was carried out in a period of time of 0.6s, so the average speed of the ant was:

v = (0 cm; 0 cm) / 0.6s = (0; 0) cm / s.

Conclusion: average speed 20 cm / s, but the average speed is zero in the ABCA path.

Examples with uniform speed on curved sections

Example 3

An insect moves on a circle with a radius of 0.2 m with uniform speed, such that starting from A and arriving at B, it travels ¼ of a circumference in 0.25 s.

Figure 4. Insect in circular section. Source: F. Zapata.

Determine the speed and velocity of the insect in section AB.

Solution

The length of the arc of circumference between A and B is:

Δs = 2πR / 4 = 2π (0.2m) / 4 = 0.32 m.

Applying the definition of average speed we have:

v = Δs / Δt = 0.32 m / 0.25 s = 1.28 m / s.

To calculate the average speed, it is necessary to calculate the displacement vector between the initial position A and the final position B:

Δ r = (0, R) - (R, 0) = (-R, R) = (-0.2, 0.2) m

Applying the definition of average speed, we obtain:

v = Δ r / Δt = (-0.2, 0.2) m / 0.25s = (-0.8, 0.8) m / s.

The previous expression is the average speed between A and B expressed in Cartesian form. Alternatively the average speed can be expressed in polar form, that is, module and direction:

- v - = ((-0.8) ^ 2 + 0.8 ^ 2) ^ (½) = 1.13 m / s

Direction = arctan (0.8 / (-0.8)) = arctan (-1) = -45º + 180º = 135º with respect to the X axis.

Finally, the mean velocity vector in polar form is: v = (1.13 m / s ¦ 135º).

Example 4

Assuming that the starting time of the insect in the previous example is 0s from point A, we have that its position vector at any instant t is given by:

r (t) =.

Determine the velocity and instantaneous speed for any time t.

Solution

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