# Average speed: formulas, how to calculate and solve exercises

The ** average velocity** for a moving particle is defined as the ratio between the variation in the position it experiences and the time interval used for the change. The simplest situation is where the particle moves along a straight line represented by the X axis.

Suppose the moving object occupies positions x _{1 and} x _{2} at times t _{1} and _{2} respectively. The definition of average velocity *v*_{ m} is mathematically represented as follows:

The units of *v*_{ m} in the International System are meters / second (m / s). Other commonly used units that appear on text and mobile devices are: km / h, cm / s, miles / h, feet / s and more, as long as they have the form duration / time.

The Greek letter “Δ” is read as “delta” and is used to briefly indicate the difference between two quantities.

__Average velocity characteristics of vector v ___{m}

__Average velocity characteristics of vector v__

_{m}

The average velocity is a vector, as it is related to the change in position, which in turn is known as the *displacement vector* .

This quality is represented in bold or by an arrow above the letter designating the magnitude. However, in a dimension, the only possible direction is the x-axis, so vector notation can be dispensed with.

Since vectors have magnitude, direction, and direction, an initial analysis of the equation indicates that the average velocity will have the same direction and direction as the displacement.

Imagine the example particle moving along a straight line. To describe its movement, it is necessary to indicate a reference point, which will be the “origin” and will be indicated as O.

The particle can move away from or towards O, to the left or to the right. It can also take a long or a short time to reach a certain position.

The mentioned magnitudes: position, displacement, time interval and average velocity describe the behavior of the particle as it moves. It’s about *kinematic* magnitudes .

To distinguish positions or locations to the left of O, the (-) sign is used and those to the right of O bear the (+) sign.

The average velocity has a geometric interpretation that can be seen in the figure below. It is the slope of the line passing through points P and Q. When cutting at position vs. curve. time in two points, is a *secant* line .

Geometric interpretation of mean velocity, as a slope of the straight line joining points P and Q. Source: す す に シ CC0 [CC0].

__Signals of average speed__

__Signals of average speed__

For the following analysis, it should be taken into account that *t *_{2}* > t ** _{1}* . In other words, the next moment is always greater than the current one. Thus,

*t*

_{2}*– t*

*is always positive, which usually makes sense on a daily basis.*

_{1}Then the sign of the mean velocity will be determined by that of *x *_{2}* – x ** _{1}* . Note that it is important to clarify where the point O – the origin – is, as this is the point at which the particle is said to go “to the right” or “to the left”.

“Forward” or “Back” as the reader prefers.

If the average velocity is positive, it means that, *on average,* the value of ” *x* ” increases with time, although this does not mean that it may have decreased at some point in the considered period – *Δt* -.

However, overall, at the end of the *Δt* times , she ended up with a bigger position than she had at the beginning. Movement details are ignored in this analysis.

**Example 1** : Given the indicated start and end positions, indicate the mean speed sign. Where did the particle move globally?

a) x _{1} = 3 m; x _{2} = 8 m

**Answer** : x _{2 }*– x _{1}* = 8 m – 3 m = 5 m. Average positive velocity, the particle advanced.

b) x _{1} = 2 m; x _{2} = -3 m

**Answer** : *x *_{2}* – x ** _{1}* = -3 m – 2 m = -5 m. Average negative velocity, the particle has moved backwards.

c) x _{1} = – 5 m; x _{2} = -12 m

**Answer** : x _{2 }*– x _{1}* = -12 m – (-5 m) = -7 m. Average negative velocity, the particle has moved backwards.

d) x _{1} = – 4 m; x _{2} = 10 m

**Answer** : x _{2} – x _{1} = 10 m – (-4 m) = 14 m. Average positive velocity, the particle advanced.

Can the average speed be 0? Yes. As long as the starting point and the ending point are the same. Does this mean that the particle was necessarily at rest all along?

No, that just means the trip was round trip. Maybe he traveled too fast or too slow. For the time being it is not known.

__Average speed: a scalar magnitude__

__Average speed: a scalar magnitude__

This leads us to define a new term: *average speed* . In physics, it is important to distinguish between vector and non-vector magnitudes: the scalars.

For the particle that made the round trip, the average speed is 0, but it could have been very fast or maybe not. To know it, the average speed is defined as:

Average speed units are the same as average speed units. The fundamental difference between the two magnitudes is that the average velocity includes interesting information about the particle’s direction and direction.

On the other hand, average speed only provides numerical information. With it, you know how fast or slow the particle has moved, but it has not moved forward or backward. That’s why it’s a scalar magnitude. How to distinguish them, denoting them? One way is to make the letters bold for the vectors or put a small arrow on them.

And it’s important to note that average speed doesn’t have to equal average speed. For the round trip, the average speed is zero, but the average speed is not. Both have the same numerical value when traveling always in the same direction.

__Exercise solved__

__Exercise solved__

You return home from school in silence at 95 km/h by 130 km. It starts to rain and reduces speed to 65 km/h. He finally makes it home after driving for 3 hours and 20 minutes.

a) How far is your house from the school?

b) What was the average speed?

**Answers:**

a) Some previous calculations are needed:

The trip is divided into two parts, the total distance is:

d = d 1 + d _{2} , with d 1 = 130 km

t 2 = 3.33 – 1.37 hours = 1.96 hours

**Calculation of d _{2:}**

d _{2} = 65 km / hx 1.96 h = 125.4 km.

The school is d 1 + d _{2} = 255.4 km from the house.