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In this article, you will learn what we mean by **average velocity** when describing the motion of a particle.

We will go through the definition and formula for average velocity and show examples so you can see how to use the formula in practice.

We will also cover many other things that you should know about average velocity like how to find the average velocity from a position vs time graph.

Let's consider a particle that is moving along a straight line:

Let's place the x-axis on the straight line along which the particle moves, with an origin point and a positive direction of our choice.

This gives us a reference relative to which we can specify the position of the particle at any point in time.

Let's also use a clock so that we know the time at which the particle has a particular position on the x-axis.

As the particle moves, with this setup in place, we see the time at which it has a particular position and we can specify what that position is relative to the x-axis.

Let's say that at some instant t_{1}, the particle is at position x_{1}, and at some subsequent instant t_{2}, the particle is at position x_{2}.

The change in position that occurs between t_{1} and t_{2} is equal to x_{2} − x_{1}. This is often indicated with Δx.

x_{2} − x_{1} = Δx

The interval of time between t_{1} and t_{2} is equal to t_{2} − t_{1}. This is often indicated with Δt:

t_{2} − t_{1} = Δt

So, how do we define the average velocity of the particle?

The average velocity of the particle between an instant t_{1} and a subsequent instant t_{2} is equal to the change in position that occurs in the interval of time between t_{1} and t_{2} divided by the interval of time itself. Therefore, if we use the symbol v to indicate the average velocity, the **formula for the average velocity** is

v = | x_{2} − x_{1} | = | Δx |

t_{2} − t_{1} | Δt |

In other words, the average velocity v for an interval of time Δt is equal to the change in position Δx that occurs during that interval of time divided by the interval of time itself:

v = | Δx |

Δt |

Therefore, what the average velocity measures is the **average rate of change of position with respect to time**.

Since average velocity is the ratio of a length to a time, the SI unit of average velocity is **meters per second** (m/s).

Another unit that is commonly used to express average velocities is **kilometers per hour** (km/h).

To convert from meters per second to kilometers per hour, simply multiply the number by 3.6. This is because:

m | = | 10^{−3} km | = 10^{−3} × 3600 | km | = 3.6 | km |

s | 1/3600 h | h | h |

For example, 20 m/s converted to km/h is

20 × 3.6 km/h = 72 km/h

Conversely, to convert from kilometers per hour to meters per second, simply divide the number by 3.6.

For example, 90 km/h converted to m/s is

90 | m/s = 25 m/s |

3.6 |

A rifle fires a bullet that travels 100 m and then hits a wall. The time that passes between the moment the bullet is fired and the moment it hits the wall is 0.250 s. What is the average velocity of the bullet for the entire trip?

Let's draw the x-axis along which the bullet moves with the positive direction coinciding with the direction of motion of the bullet:

Let's assume that the instant the bullet is fired is 0 and label it as t_{1}. Also, let's consider the origin of the x-axis to be right where the bullet is at the moment it is fired. So, the position x_{1} at instant t_{1} is 0.

t_{1} = 0, x_{1} = 0

We know that at a subsequent instant t_{2}, the bullet hits the wall after having traveled 100 m. So, the position x_{2} at instant t_{2} is 100 m. Also, since the time that passes between the instant t_{1} at which the bullet is fired and the instant t_{2} at which it hits the wall is 0.250 s, it follows that t_{2} is 0.250 s.

t_{2} = 0.250 s, x_{2} = 100 m

The entire trip of the bullet starts at t_{1} and ends at t_{2}. So, the average velocity for the entire trip is the average velocity between t_{1} and t_{2}, which we indicate with v_{12}.

As we've seen, the average velocity v_{12} between an instant t_{1} and an instant t_{2} is equal to the ratio of the change in position that happens between t_{1} and t_{2} to the interval of time between t_{1} and t_{2}:

v_{12} = | x_{2} − x_{1} | = | 100 m − 0 | = | 100 m | = 400 m/s |

t_{2} − t_{1} | 0.250 s − 0 | 0.250 s |

Thus, the average velocity of the bullet for the entire trip is 400 m/s.

A car travels 120 km in 1.5 h (one and a half hour). After that, it travels 60 km in 0.50 h (half an hour), arriving at its final destination. Calculate the average velocity for the first part of the trip, the average velocity for the second part of the trip, and the average velocity for the entire trip.

We start by representing the x-axis along which the car moves with the positive direction coinciding with the direction of motion of the car:

Let's suppose that the instant at which the car starts moving is 0 and label it as t_{1}. Also, let's consider the origin of the x-axis to coincide with the position of the car at the moment it starts moving. So, the position x_{1} at instant t_{1} is 0.

t_{1} = 0, x_{1} = 0

At a subsequent instant t_{2}, the car has traveled 120 km. So, the position x_{2} at instant t_{2} is 120 km. The time that passes between the instant t_{1} at which the car starts and the instant t_{2} at which it has traveled 120 km is 1.5 h, so t_{2} is 1.5 h.

t_{2} = 1.5 h, x_{2} = 120 km

Later, at instant t_{3}, the car has traveled another 60 km in 0.50 h. So, at instant t_{3}, the car has a position x_{3} which is greater than x_{2}.

We can write

x_{3} − x_{2} = 60 km

t_{3} − t_{2} = 0.50 h

We need to calculate the average velocity for the first part of the trip, which starts at t_{1} and ends at t_{2}, the second part of the trip, which starts at t_{2} and ends at t_{3}, and the entire trip, which starts at t_{1} and ends at t_{3}.

Let's begin by finding the average velocity v_{12} between t_{1} and t_{2}:

v_{12} = | x_{2} − x_{1} | = | 120 km − 0 | = | 120 km | = 80 km/h |

t_{2} − t_{1} | 1.5 h − 0 | 1.5 h |

Next, let's find the average velocity v_{23} between t_{2} and t_{3}:

v_{23} = | x_{3} − x_{2} | = | 60 km | = 120 km/h |

t_{3} − t_{2} | 0.50 h |

Finally, let's calculate the average velocity v_{13} between t_{1} and t_{3}:

v_{13} = | x_{3} − x_{1} |

t_{3} − t_{1} |

By looking at the x-axis with the positions of the car at different times, we can see that the change in position x_{3} − x_{1} can be written as the sum of x_{2} − x_{1} and x_{3} − x_{2}:

x_{3} − x_{1} = (x_{2} − x_{1}) + (x_{3} − x_{2})

x_{3} − x_{1} = 120 km + 60 km

x_{3} − x_{1} = 180 km

Similarly, the interval of time t_{3} − t_{1} can be written as the sum of t_{2} − t_{1} and t_{3} − t_{2}:

t_{3} − t_{1} = (t_{2} − t_{1}) + (t_{3} − t_{2})

t_{3} − t_{1} = 1.5 h + 0.50 h

t_{3} − t_{1} = 2.0 h

Thus, the average velocity v_{13} for the entire trip is

v_{13} = | 180 km | = 90 km/h |

2.0 h |

Notice that the average velocity v_{13} for the entire trip is **not** necessarily equal to the arithmetic mean of the average velocity v_{12} for the first part of the trip and the average velocity v_{23} for the second part of the trip. Let's show this by actually calculating the arithmetic mean of v_{12} and v_{23}:

v_{12} + v_{23} | = | 80 km/h + 120 km/h | = 100 km/h |

2 | 2 |

As you can see, the arithmetic mean is different from the actual average velocity for the entire trip.

Therefore, it is important that you do not assume that the average velocity for the whole is the arithmetic mean of the average velocities for the parts.

Since the average velocity is the change in position divided by the interval of time

v = | Δx |

Δt |

and the interval of time Δt is always positive, it follows that the sign of the average velocity v is the same as the sign of the change in position Δx.

The change in position Δx can be positive, negative, or zero, depending on whether the particle moved in the positive x-direction, negative x-direction, or ended up at the same position from which it started. Let's see these 3 cases in more detail:

- Δx > 0
- Δx < 0
- Δx = 0

Therefore, the average velocity v can be positive, negative, or zero:

- Δx > 0 → v > 0
- Δx < 0 → v < 0
- Δx = 0 → v = 0

A train moves from one station to another in 30 minutes. Then, after staying there for 15 minutes, it returns back to the original station in 30 minutes. Knowing that the distance between the two stations is 35 km, find the average velocity of the train for the trip from the first station to the second station, the average velocity for the trip back, and the average velocity for the round trip.

As usual, let's represent the x-axis along which the train moves having the positive direction in the direction of motion of the train:

We will consider the instant at which the train starts moving to be 0 and we will refer to it as t_{1}. We will also consider the origin of the x-axis to coincide with the position that the train has at the moment it begins moving. Thus, the position x_{1} at instant t_{1} is 0.

t_{1} = 0, x_{1} = 0

Then, at an instant t_{2}, the train arrives at the other station after having traveled 35 km. Therefore, the position x_{2} at instant t_{2} is equal to 35 km. The amount of time between the instant t_{1} and the instant t_{2} is 30 minutes, which in hours is 0.50 h, so t_{2} is equal to 0.50 h.

t_{2} = 0.50 h, x_{2} = 35 km

The train remains at the second station for 15 minutes, i.e. 0.25 h, and then heads back to the first station. Let's refer to the instant at which the train starts moving back as t_{3}. We know that t_{3} is 0.25 h after t_{2} (the instant at which it arrived at the second station). The position x_{3} that the train has at t_{3} is the same as the position at t_{2}, i.e. 35 km.

t_{3} − t_{2} = 0.25 h

x_{3} = 35 km

Finally, the train returns to the first station in 30 minutes, i.e. 0.50 h. Let's refer to the instant at which the train is back as t_{4}. This instant t_{4} is 0.50 h after the instant t_{3} at which the train departed from the second station. The position x_{4} that the train has when it's back is equal to 0.

t_{4} − t_{3} = 0.50 h

x_{4} = 0

The problem wants us to find the average velocity for the trip from the first station to the second station, which starts at t_{1} and ends at t_{2}, the trip back, which starts at t_{3} and ends at t_{4}, and the round trip, which starts at t_{1} and ends at t_{4}.

Let's start by calculating the average velocity v_{12} between t_{1} and t_{2}:

v_{12} = | x_{2} − x_{1} | = | 35 km − 0 | = | 35 km | = 70 km/h |

t_{2} − t_{1} | 0.50 h − 0 | 0.50 h |

Next, let's calculate the average velocity v_{34} between t_{3} and t_{4}:

v_{34} = | x_{4} − x_{3} | = | 0 − 35 km | = | −35 km | = −70 km/h |

t_{4} − t_{3} | 0.50 h | 0.50 h |

Lastly, let's find the average velocity v_{14} between t_{1} and t_{4}:

v_{14} = | x_{4} − x_{1} | = | 0 − 0 | = | 0 | = 0 |

t_{4} − t_{1} | t_{4} − t_{1} | t_{4} − t_{1} |

Therefore, the train has a positive average velocity (70 km/h) for the trip from the first station to the second station, a negative average velocity (−70 km/h) for the trip from the second station back to the first station, and a zero average velocity for the round trip.

It is common to describe the motion of a particle using a **position vs time graph**:

A position vs time graph shows how the position of a particle changes over time.

We can use a position vs time graph to determine the position of a particle at any instant. For example, in the previous position-time graph, at the instant t = 3 s, the particle is at position x = 20 m:

This means that the position vs time graph gives us all the information that we need in order to find the average velocity of the particle for any interval of time. For example, let's say that we want to find the average velocity between 3 s and 5 s from the previous position-time graph. We can do this by following these 3 simple steps:

- Look at the graph and determine the position x
_{1}that the particle has at instant t_{1}= 3 s. - Look at the graph and determine the position x
_{2}that the particle has at instant t_{2}= 5 s. - Calculate the average velocity by dividing the change in position x
_{2}− x_{1}by the interval of time t_{2}− t_{1}.

Let's begin with steps 1 and 2:

t_{1} = 3 s, x_{1} = 20 m

t_{2} = 5 s, x_{2} = 30 m

In step 3, we simply calculate the average velocity:

v_{12} = | x_{2} − x_{1} | = | 30 m − 20 m | = | 10 m | = 5 m/s |

t_{2} − t_{1} | 5 s − 3 s | 2 s |

Let's say that we have the position-time graph for the motion of a particle:

Let's consider an instant t_{1} at which the particle has position x_{1} and a subsequent instant t_{2} at which the particle has position x_{2}:

Notice that the change in position x_{2} − x_{1} = Δx is the vertical distance between the points corresponding to t_{1} and t_{2} on the graph:

and the interval of time t_{2} − t_{1} = Δt is the horizontal distance between those points:

Let's draw a straight line that passes through the two points (a secant line):

We know that the average velocity for Δt is the ratio of Δx to Δt:

v = | Δx |

Δt |

But as we can see from the graph above, the ratio of Δx to Δt is at the same time the ratio of the vertical change to the horizontal change between two points on the secant line. That is, the ratio of Δx to Δt is equal to the slope of the secant line. If we call m the slope of the secant line, we can write:

Δx | = m |

Δt |

So, the slope m of the secant line is equal to the average velocity v:

m = v

Therefore, the average velocity between an instant t_{1} and an instant t_{2} is equal to the slope of the secant line that passes through the points t_{1} and t_{2} on the position vs time graph.

Let's return to the position-time graph that we've seen in the previous section, again considering an instant t_{1} at which the particle has position x_{1} and a subsequent instant t_{2} at which the particle has position x_{2}:

Let's call θ the angle that the secant line passing through the points t_{1} and t_{2} on the position vs time graph makes with the positive t-axis:

We can look at the sign of the angle θ to determine whether the average velocity is positive, negative, or zero:

- θ > 0Δx > 0 → v > 0
- θ < 0Δx < 0 → v < 0
- θ = 0Δx = 0 → v = 0

We know that the average velocity v for an interval of time Δt is equal to the change in position Δx that occurs during Δt divided by Δt itself:

v = | Δx |

Δt |

We can use this equation to find the change in position Δx:

Δx = vΔt

This tells us that the change in position Δx that occurs in an interval of time Δt is equal to the average velocity v for that interval of time multiplied by the interval of time itself.

We can also use the above equation to find the interval of time Δt:

Δt = | Δx |

v |

This tells us that an interval of time Δt is equal to the change in position Δx that occurs during that interval of time divided by the average velocity v for that interval of time.

In a marathon, a runner runs the first 26.2 km with an average velocity of 3.52 m/s. Then, he finishes the marathon after covering the remaining distance in 1.56 h with an average velocity of 2.85 m/s. Calculate the average velocity of the runner for the entire race.

Let's represent the x-axis along which the runner runs, with the positive direction coinciding with the direction of motion of the runner:

Let's suppose that the instant at which the race starts is 0 and let's label it as t_{1}. Also, let's consider the origin of the x-axis to be right at the point where the runner is when the race starts. So, the position x_{1} at instant t_{1} is 0.

t_{1} = 0, x_{1} = 0

At a subsequent instant t_{2}, the runner has covered the first 26.2 km. So, the position x_{2} at instant t_{2} is 26.2 km. The instant t_{2} is not given. However, we are given the average velocity between t_{1} and t_{2}, which is 3.52 m/s.

t_{2} = unknown, x_{2} = 26.2 km

v_{12} = 3.52 m/s

Finally, at instant t_{3}, the runner reaches a position x_{3} which corresponds to the finish line. The instant t_{3} and the position x_{3} are not specified in the problem. However, we know the interval of time between t_{2} and t_{3}, which is 1.56 h, and the average velocity between t_{2} and t_{3}, which is 2.85 m/s.

t_{3} = unknown, x_{3} = unknown

t_{3} − t_{2} = 1.56 h

v_{23} = 2.85 m/s

We need to calculate the average velocity for the entire race, which starts at t_{1} and ends at t_{3}.

The average velocity v_{13} between t_{1} and t_{3} is

v_{13} = | x_{3} − x_{1} |

t_{3} − t_{1} |

So, in order to calculate v_{13}, we first need to determine what x_{3} − x_{1} is and what t_{3} − t_{1} is.

As we've seen in the car example, the change in position x_{3} − x_{1} is equal to the sum of x_{2} − x_{1} and x_{3} − x_{2}:

x_{3} − x_{1} = (x_{2} − x_{1}) + (x_{3} − x_{2})

x_{2} − x_{1} can be easily calculated since we know both x_{1} and x_{2}:

x_{2} − x_{1} = 26.2 km − 0

x_{2} − x_{1} = 26.2 km

But what about x_{3} − x_{2}? We know x_{2} but not x_{3}. However, we do know the interval of time between t_{2} and t_{3} as well as the average velocity between t_{2} and t_{3}:

t_{3} − t_{2} = 1.56 h

v_{23} = 2.85 m/s

And knowing that

v_{23} = | x_{3} − x_{2} |

t_{3} − t_{2} |

We can solve this for x_{3} − x_{2}:

x_{3} − x_{2} = v_{23} (t_{3} − t_{2})

x_{3} − x_{2} = (2.85 m/s) (1.56 h)

Remember that to convert from m/s to km/h, we multiply the number by 3.6:

x_{3} − x_{2} = (2.85 × 3.6 km/h) (1.56 h)

x_{3} − x_{2} = 16.0 km

Having found x_{3} − x_{2}, we can now determine x_{3} − x_{1}:

x_{3} − x_{1} = (x_{2} − x_{1}) + (x_{3} − x_{2})

x_{3} − x_{1} = 26.2 km + 16.0 km

x_{3} − x_{1} = 42.2 km

Next, we need to find t_{3} − t_{1}.

t_{3} − t_{1} is the sum of t_{2} − t_{1} and t_{3} − t_{2}:

t_{3} − t_{1} = (t_{2} − t_{1}) + (t_{3} − t_{2})

t_{3} − t_{2} is already known:

t_{3} − t_{2} = 1.56 h

But what about t_{2} − t_{1}? We know t_{1} but not t_{2}. However, we know the change in position between t_{1} and t_{2} as well as the average velocity between t_{1} and t_{2}:

x_{2} − x_{1} = 26.2 km

v_{12} = 3.52 m/s

And since

v_{12} = | x_{2} − x_{1} |

t_{2} − t_{1} |

We can solve this for t_{2} − t_{1}:

(t_{2} − t_{1}) v_{12} = x_{2} − x_{1}

t_{2} − t_{1} = | x_{2} − x_{1} |

v_{12} |

t_{2} − t_{1} = | 26.2 km |

3.52 m/s |

t_{2} − t_{1} = | 26.2 km |

3.52 × 3.6 km/h |

t_{2} − t_{1} = 2.07 h

We can now determine t_{3} − t_{1}:

t_{3} − t_{1} = (t_{2} − t_{1}) + (t_{3} − t_{2})

t_{3} − t_{1} = 2.07 h + 1.56 h

t_{3} − t_{1} = 3.63 h

Finally, we can find the average velocity v_{13} for the entire race:

v_{13} = | x_{3} − x_{1} |

t_{3} − t_{1} |

v_{13} = | 42.2 km | = 11.6 km/h |

3.63 h |

Let's also find the average velocity v_{13} expressed in m/s.

Remembering that to convert from km/h to m/s, we divide the number by 3.6, we can write:

v_{13} = | 11.6 | m/s = 3.22 m/s |

3.6 |

Therefore, the average velocity of the runner for the entire race is 11.6 km/h, or equivalently 3.22 m/s.

The average velocity v between an instant t

_{1}and an instant t_{2}is the ratio of the change in position x_{2}− x_{1}= Δx to the interval of time t_{2}− t_{1}= Δt:v = x _{2}− x_{1}= Δx t _{2}− t_{1}Δt The SI unit of average velocity is meters per second (m/s). Another unit that is commonly used is kilometers per hour (km/h). To convert from m/s to km/h, multiply by 3.6. To convert from km/h to m/s, divide by 3.6.

The sign of the average velocity v is the same as the sign of the change in position Δx.

- Δx > 0 → v > 0
- Δx < 0 → v < 0
- Δx = 0 → v = 0

In the context of a position vs time graph, the average velocity between an instant t

_{1}and an instant t_{2}is equal to the slope of the secant line that passes through the points t_{1}and t_{2}on the graph.You can determine the sign of the average velocity by looking at the sign of the angle θ that the secant line makes with the positive t-axis.

- θ > 0 → v > 0
- θ < 0 → v < 0
- θ = 0 → v = 0

The change in position Δx that occurs in an interval of time Δt is equal to the average velocity v for that interval of time multiplied by the interval of time itself:

Δx = vΔtAn interval of time Δt is equal to the change in position Δx that occurs during that interval of time divided by the average velocity v for that interval of time:

Δt = Δx v

A professional athlete finishes a 100 m run in 12.3 s. Calculate the average velocity of the athlete for the entire run.

8.13 m/s

t_{1} = 0, x_{1} = 0

t_{2} = 12.3 s, x_{2} = 100 m

v_{12} = ?

v_{12} = | x_{2} − x_{1} | = | 100 m − 0 | = | 100 m | = 8.13 m/s |

t_{2} − t_{1} | 12.3 s − 0 | 12.3 s |

A person walks 250 m, from his house to the gym, in 155 s. Then, he gets to the park, which is 320 m away from the gym, in 272 s. Assuming that all the walking happens in the same direction, what are the average velocities for the walk from the house to the gym, from the gym to the park, and from the house to the park?

1.61 m/s; 1.18 m/s; 1.33 m/s

t_{1} = 0, x_{1} = 0

t_{2} = 155 s, x_{2} = 250 m

t_{3} − t_{2} = 272 s, x_{3} − x_{2} = 320 m

v_{12} = ?, v_{23} = ?, v_{13} = ?

v_{12} = | x_{2} − x_{1} | = | 250 m − 0 | = | 250 m | = 1.61 m/s |

t_{2} − t_{1} | 155 s − 0 | 155 s |

v_{23} = | x_{3} − x_{2} | = | 320 m | = 1.18 m/s |

t_{3} − t_{2} | 272 s |

v_{13} = | x_{3} − x_{1} |

t_{3} − t_{1} |

x_{3} − x_{1} = (x_{2} − x_{1}) + (x_{3} − x_{2})

x_{3} − x_{1} = 250 m + 320 m

x_{3} − x_{1} = 570 m

t_{3} − t_{1} = (t_{2} − t_{1}) + (t_{3} − t_{2})

t_{3} − t_{1} = 155 s + 272 s

t_{3} − t_{1} = 427 s

v_{13} = | x_{3} − x_{1} | = | 570 m | = 1.33 m/s |

t_{3} − t_{1} | 427 s |

A bus travels from one city to another in 50 minutes. When the bus arrives at the destination city, it remains there for 30 minutes. Then, it returns back to the original city in 45 minutes. The road distance between the two cities is 60 km. Find the average velocities that the bus has for the trip from the first city to the second city, for the trip back, and for the round trip.

72 km/h; −80 km/h; 0

t_{1} = 0, x_{1} = 0

t_{2} = 50 min, x_{2} = 60 km

t_{3} − t_{2} = 30 min, x_{3} = 60 km

t_{4} − t_{3} = 45 min, x_{4} = 0

v_{12} = ?, v_{34} = ?, v_{14} = ?

v_{12} = | x_{2} − x_{1} | = | 60 km − 0 | = | 60 km |

t_{2} − t_{1} | 50 min − 0 | 50 min |

50 min = 50 (1/60 h) = 0.83 h

v_{12} = | 60 km | = 72 km/h |

0.83 h |

v_{34} = | x_{4} − x_{3} | = | 0 − 60 km | = | −60 km |

t_{4} − t_{3} | 45 min | 45 min |

45 min = 45 (1/60 h) = 0.75 h

v_{34} = | −60 km | = −80 km/h |

0.75 h |

v_{14} = | x_{4} − x_{1} | = | 0 − 0 | = | 0 | = 0 |

t_{4} − t_{1} | t_{4} − t_{1} | t_{4} − t_{1} |

In a bicycle race, a rider has an average velocity of 8.83 m/s for the first 2.15 h. Then, the rider crosses the finish line after covering the remaining 43.7 km with an average velocity of 7.35 m/s. What is the average velocity of the rider for the entire race expressed both in m/s and in km/h?

8.19 m/s; 29.5 km/h

t_{1} = 0, x_{1} = 0

t_{2} = 2.15 h, x_{2} = unknown, v_{12} = 8.83 m/s

t_{3} = unknown, x_{3} − x_{2} = 43.7 km, v_{23} = 7.35 m/s

v_{13} = ?

v_{13} = | x_{3} − x_{1} |

t_{3} − t_{1} |

x_{3} − x_{1} = (x_{2} − x_{1}) + (x_{3} − x_{2})

x_{3} − x_{2} = 43.7 km

t_{2} − t_{1} = 2.15 h − 0 = 2.15 h

v_{12} = 8.83 m/s

v_{12} = | x_{2} − x_{1} |

t_{2} − t_{1} |

x_{2} − x_{1} = v_{12} (t_{2} − t_{1})

x_{2} − x_{1} = (8.83 m/s) (2.15 h)

x_{2} − x_{1} = (8.83 × 3.6 km/h) (2.15 h)

x_{2} − x_{1} = 68.3 km

x_{3} − x_{1} = (x_{2} − x_{1}) + (x_{3} − x_{2})

x_{3} − x_{1} = 68.3 km + 43.7 km

x_{3} − x_{1} = 112.0 km

t_{3} − t_{1} = (t_{2} − t_{1}) + (t_{3} − t_{2})

t_{2} − t_{1} = 2.15 h

x_{3} − x_{2} = 43.7 km

v_{23} = 7.35 m/s

v_{23} = | x_{3} − x_{2} |

t_{3} − t_{2} |

(t_{3} − t_{2}) v_{23} = x_{3} − x_{2}

t_{3} − t_{2} = | x_{3} − x_{2} |

v_{23} |

t_{3} − t_{2} = | 43.7 km |

7.35 m/s |

t_{3} − t_{2} = | 43.7 km |

7.35 × 3.6 km/h |

t_{3} − t_{2} = 1.65 h

t_{3} − t_{1} = (t_{2} − t_{1}) + (t_{3} − t_{2})

t_{3} − t_{1} = 2.15 h + 1.65 h

t_{3} − t_{1} = 3.80 h

v_{13} = | x_{3} − x_{1} |

t_{3} − t_{1} |

v_{13} = | 112.0 km | = 29.5 km/h |

3.80 h |

v_{13} = | 29.5 | m/s = 8.19 m/s |

3.6 |