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Average Velocity: Definition, Formula, Examples and more

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.

Definition and formula for average velocity

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

A particle on a straight line.ParticleLine on which it moves

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.

The particle on the x-axis.xO

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 t1, the particle is at position x1, and at some subsequent instant t2, the particle is at position x2.

The change in position that occurs between t1 and t2 is equal to x2 − x1. This is often indicated with Δx.

x-axis with an indication of the change in position from x1 to x2.xO1x2xΔx1t2t
x2x1 = Δx

The interval of time between t1 and t2 is equal to t2 − t1. This is often indicated with Δt:

t2t1 = Δt

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

The average velocity of the particle between an instant t1 and a subsequent instant t2 is equal to the change in position that occurs in the interval of time between t1 and t2 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

vx2x1 = Δx
t2t1Δt

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:

vΔx
Δt

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

Unit of average velocity

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
s1/3600 hhh

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

Example: Average velocity of a bullet

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:

x-axis with positive direction going to the right.xO

Let's assume that the instant the bullet is fired is 0 and label it as t1. 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 x1 at instant t1 is 0.

t1 = 0, x1 = 0
x-axis with the position x1 at the origin.xO1x

We know that at a subsequent instant t2, the bullet hits the wall after having traveled 100 m. So, the position x2 at instant t2 is 100 m. Also, since the time that passes between the instant t1 at which the bullet is fired and the instant t2 at which it hits the wall is 0.250 s, it follows that t2 is 0.250 s.

t2 = 0.250 s, x2 = 100 m
x-axis with the position x1 at the origin and the position x2 positive.xO1x2x

The entire trip of the bullet starts at t1 and ends at t2. So, the average velocity for the entire trip is the average velocity between t1 and t2, which we indicate with v12.

As we've seen, the average velocity v12 between an instant t1 and an instant t2 is equal to the ratio of the change in position that happens between t1 and t2 to the interval of time between t1 and t2:

v12x2x1 = 100 m0 = 100 m = 400 m/s
t2t10.250 s00.250 s

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

Example: Average velocity of a car

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:

x-axis with positive direction going to the right.xO

Let's suppose that the instant at which the car starts moving is 0 and label it as t1. 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 x1 at instant t1 is 0.

t1 = 0, x1 = 0
x-axis with the position x1 at the origin.xO1x

At a subsequent instant t2, the car has traveled 120 km. So, the position x2 at instant t2 is 120 km. The time that passes between the instant t1 at which the car starts and the instant t2 at which it has traveled 120 km is 1.5 h, so t2 is 1.5 h.

t2 = 1.5 h, x2 = 120 km
x-axis with position x1 at the origin and position x2 positive.xO1x2x

Later, at instant t3, the car has traveled another 60 km in 0.50 h. So, at instant t3, the car has a position x3 which is greater than x2.

x-axis with position x1 at the origin, position x2 positive, and position x3 greater than x2.xO1x2x3x

We can write

x3x2 = 60 km
t3t2 = 0.50 h

We need to calculate the average velocity for the first part of the trip, which starts at t1 and ends at t2, the second part of the trip, which starts at t2 and ends at t3, and the entire trip, which starts at t1 and ends at t3.

Let's begin by finding the average velocity v12 between t1 and t2:

v12x2x1 = 120 km0 = 120 km = 80 km/h
t2t11.5 h01.5 h

Next, let's find the average velocity v23 between t2 and t3:

v23x3x2 = 60 km = 120 km/h
t3t20.50 h

Finally, let's calculate the average velocity v13 between t1 and t3:

v13x3x1
t3t1

By looking at the x-axis with the positions of the car at different times, we can see that the change in position x3 − x1 can be written as the sum of x2 − x1 and x3 − x2:

x-axis with the change in position from x1 to x3 seen as the sum of the change in position from x1 to x2 and from x2 to x3.xO1x2x3x3x1x2x1x3x2x
x3x1 = (x2x1) + (x3x2)
x3x1 = 120 km + 60 km
x3x1 = 180 km

Similarly, the interval of time t3 − t1 can be written as the sum of t2 − t1 and t3 − t2:

t3t1 = (t2t1) + (t3t2)
t3t1 = 1.5 h + 0.50 h
t3t1 = 2.0 h

Thus, the average velocity v13 for the entire trip is

v13180 km = 90 km/h
2.0 h

Notice that the average velocity v13 for the entire trip is not necessarily equal to the arithmetic mean of the average velocity v12 for the first part of the trip and the average velocity v23 for the second part of the trip. Let's show this by actually calculating the arithmetic mean of v12 and v23:

v12 + v23 = 80 km/h + 120 km/h = 100 km/h
22

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.

Sign of average velocity

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

vΔx
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:

  1. Δx > 0
    x-axis with position x2 larger than position x1.xO1x2xΔx
  2. Δx < 0
    x-axis with position x2 smaller than position x1.xO2x1xΔx
  3. Δx = 0
    x-axis with position x2 equal to position x1.xO1x2x=

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

  1. Δx > 0 v > 0
  2. Δx < 0 v < 0
  3. Δx = 0 v = 0

Example: Average velocity of a train

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:

x-axis with positive direction going to the right.xO

We will consider the instant at which the train starts moving to be 0 and we will refer to it as t1. 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 x1 at instant t1 is 0.

t1 = 0, x1 = 0
x-axis with the position x1 at the origin.xO1x

Then, at an instant t2, the train arrives at the other station after having traveled 35 km. Therefore, the position x2 at instant t2 is equal to 35 km. The amount of time between the instant t1 and the instant t2 is 30 minutes, which in hours is 0.50 h, so t2 is equal to 0.50 h.

t2 = 0.50 h, x2 = 35 km
x-axis with position x1 at the origin and position x2 positive.xO1x2x

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 t3. We know that t3 is 0.25 h after t2 (the instant at which it arrived at the second station). The position x3 that the train has at t3 is the same as the position at t2, i.e. 35 km.

t3t2 = 0.25 h
x3 = 35 km
x-axis with position x1 at the origin, position x2 positive, and position x3 equal to x2.xO1x2x3x=

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 t4. This instant t4 is 0.50 h after the instant t3 at which the train departed from the second station. The position x4 that the train has when it's back is equal to 0.

t4t3 = 0.50 h
x4 = 0
x-axis with position x1 at the origin, position x2 positive, position x3 equal to x2, and position x4 at the origin.xO2x3x=1x4x=

The problem wants us to find the average velocity for the trip from the first station to the second station, which starts at t1 and ends at t2, the trip back, which starts at t3 and ends at t4, and the round trip, which starts at t1 and ends at t4.

Let's start by calculating the average velocity v12 between t1 and t2:

v12x2x1 = 35 km0 = 35 km = 70 km/h
t2t10.50 h00.50 h

Next, let's calculate the average velocity v34 between t3 and t4:

v34x4x3 = 035 km = −35 km = −70 km/h
t4t30.50 h0.50 h

Lastly, let's find the average velocity v14 between t1 and t4:

v14x4x1 = 00 = 0 = 0
t4t1t4t1t4t1

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.

Position vs time graphs and average velocity

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

Position vs time graph where the position at time 0 is 0, then becomes positive, and finally, at time 9 seconds it returns back to 0.x(m)1050403020t(s)897654321O

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

Finding average velocity from a position vs time graph

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:

Position vs time graph in which at 3 seconds the position is 20 meters.x(m)t(s)O8976543211050403020

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:

  1. Look at the graph and determine the position x1 that the particle has at instant t1 = 3 s.
  2. Look at the graph and determine the position x2 that the particle has at instant t2 = 5 s.
  3. Calculate the average velocity by dividing the change in position x2 − x1 by the interval of time t2 − t1.

Let's begin with steps 1 and 2:

Position vs time graph in which at 3 seconds the position is 20 meters and at 5 seconds 30 meters.x(m)t(s)O1050403020897654321
t1 = 3 s, x1 = 20 m
t2 = 5 s, x2 = 30 m

In step 3, we simply calculate the average velocity:

v12x2x1 = 30 m20 m = 10 m = 5 m/s
t2t15 s3 s2 s

Average velocity as the slope of a secant line

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

Position vs time graph where the position at time 0 is 0, then becomes positive, and finally goes back to 0.xtO

Let's consider an instant t1 at which the particle has position x1 and a subsequent instant t2 at which the particle has position x2:

Position vs time graph in which at instant t1 the position is x1 and at a subsequent instant t2 it is x2.xtO1t2txx12

Notice that the change in position x2 − x1 = Δx is the vertical distance between the points corresponding to t1 and t2 on the graph:

Position vs time graph in which the change in position is indicated as the vertical distance between the points t1 and t2 on the graph.xtO1t2tΔxxx12

and the interval of time t2 − t1 = Δt is the horizontal distance between those points:

Position vs time graph in which the interval of time is indicated as the horizontal distance between the points t1 and t2 on the graph.xtO1t2tΔxΔtxx12

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

Position vs time graph with the secant line that passes through the points t1 and t2 on the graph.xtO1t2tΔxΔtxx12

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

vΔx
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 t1 and an instant t2 is equal to the slope of the secant line that passes through the points t1 and t2 on the position vs time graph.

Looking at the secant line angle to determine the sign of average velocity

Let's return to the position-time graph that we've seen in the previous section, again considering an instant t1 at which the particle has position x1 and a subsequent instant t2 at which the particle has position x2:

Position vs time graph where at t1 the position is x1, at t2 x2, and Δx, as well as Δt, is indicated.xtO1t2tΔxΔtxx12

Let's call θ the angle that the secant line passing through the points t1 and t2 on the position vs time graph makes with the positive t-axis:

Position vs time graph with the angle θ that the secant line passing through the points t1 and t2 on the graph makes with the positive t-axis.xtO1t2tΔxΔtθxx12

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

  1. θ > 0
    Position vs time graph with the secant line angle positive.xtO1t2tΔxΔtθxx12
    Δx > 0 v > 0
  2. θ < 0
    Position vs time graph with the secant line angle negative.xtOΔxΔt1t2tθxx12
    Δx < 0 v < 0
  3. θ = 0
    Position vs time graph with the secant line angle zero.xtO1t2txx12=
    Δx = 0 v = 0

Expressing the change in position or the interval of time in terms of the average velocity

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:

vΔx
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:

ΔtΔx
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.

Example: Average velocity of a runner

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:

x-axis with positive direction going to the right.xO

Let's suppose that the instant at which the race starts is 0 and let's label it as t1. 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 x1 at instant t1 is 0.

t1 = 0, x1 = 0
x-axis with the position x1 at the origin.xO1x

At a subsequent instant t2, the runner has covered the first 26.2 km. So, the position x2 at instant t2 is 26.2 km. The instant t2 is not given. However, we are given the average velocity between t1 and t2, which is 3.52 m/s.

t2 = unknown, x2 = 26.2 km
v12 = 3.52 m/s
x-axis with position x1 at the origin and position x2 positive.xO1x2x

Finally, at instant t3, the runner reaches a position x3 which corresponds to the finish line. The instant t3 and the position x3 are not specified in the problem. However, we know the interval of time between t2 and t3, which is 1.56 h, and the average velocity between t2 and t3, which is 2.85 m/s.

t3 = unknown, x3 = unknown
t3t2 = 1.56 h
v23 = 2.85 m/s
x-axis with position x1 at the origin, position x2 positive, and position x3 greater than x2.xO1x2x3x

We need to calculate the average velocity for the entire race, which starts at t1 and ends at t3.

The average velocity v13 between t1 and t3 is

v13x3x1
t3t1

So, in order to calculate v13, we first need to determine what x3 − x1 is and what t3 − t1 is.

As we've seen in the car example, the change in position x3 − x1 is equal to the sum of x2 − x1 and x3 − x2:

x-axis with the change in position from x1 to x3 seen as the sum of the change in position from x1 to x2 and from x2 to x3.xO1x2x3x3x1x2x1x3x2x
x3x1 = (x2x1) + (x3x2)

x2 − x1 can be easily calculated since we know both x1 and x2:

x2x1 = 26.2 km0
x2x1 = 26.2 km

But what about x3 − x2? We know x2 but not x3. However, we do know the interval of time between t2 and t3 as well as the average velocity between t2 and t3:

t3t2 = 1.56 h
v23 = 2.85 m/s

And knowing that

v23x3x2
t3t2

We can solve this for x3 − x2:

x3x2 = v23 (t3t2)
x3x2 = (2.85 m/s) (1.56 h)

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

x3x2 = (2.85 × 3.6 km/h) (1.56 h)
x3x2 = 16.0 km

Having found x3 − x2, we can now determine x3 − x1:

x3x1 = (x2x1) + (x3x2)
x3x1 = 26.2 km + 16.0 km
x3x1 = 42.2 km

Next, we need to find t3 − t1.

t3 − t1 is the sum of t2 − t1 and t3 − t2:

t3t1 = (t2t1) + (t3t2)

t3 − t2 is already known:

t3t2 = 1.56 h

But what about t2 − t1? We know t1 but not t2. However, we know the change in position between t1 and t2 as well as the average velocity between t1 and t2:

x2x1 = 26.2 km
v12 = 3.52 m/s

And since

v12x2x1
t2t1

We can solve this for t2 − t1:

(t2t1) v12 = x2x1
t2t1x2x1
v12
t2t126.2 km
3.52 m/s
t2t126.2 km
3.52 × 3.6 km/h
t2t1 = 2.07 h

We can now determine t3 − t1:

t3t1 = (t2t1) + (t3t2)
t3t1 = 2.07 h + 1.56 h
t3t1 = 3.63 h

Finally, we can find the average velocity v13 for the entire race:

v13x3x1
t3t1
v1342.2 km = 11.6 km/h
3.63 h

Let's also find the average velocity v13 expressed in m/s.

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

v1311.6 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.

Summary

  • The average velocity v between an instant t1 and an instant t2 is the ratio of the change in position x2 − x1 = Δx to the interval of time t2 − t1 = Δt:

    vx2x1 = Δx
    t2t1Δt
  • 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.

    1. Δx > 0 v > 0
    2. Δx < 0 v < 0
    3. Δx = 0 v = 0
  • In the context of a position vs time graph, the average velocity between an instant t1 and an instant t2 is equal to the slope of the secant line that passes through the points t1 and t2 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.

    1. θ > 0 v > 0
    2. θ < 0 v < 0
    3. 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 = vΔt
  • 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:

    ΔtΔx
    v

Exercises

#1

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

Solution

8.13 m/s

How to arrive at the solution

x-axis with the position x1 at the origin and the position x2 positive.xO1x2x
t1 = 0, x1 = 0
t2 = 12.3 s, x2 = 100 m
v12 = ?
v12x2x1 = 100 m0 = 100 m = 8.13 m/s
t2t112.3 s012.3 s

#2

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?

Solution

1.61 m/s; 1.18 m/s; 1.33 m/s

How to arrive at the solution

x-axis with position x1 at the origin, position x2 positive, and position x3 greater than x2.xO1x2x3x
t1 = 0, x1 = 0
t2 = 155 s, x2 = 250 m
t3t2 = 272 s, x3x2 = 320 m
v12 = ?, v23 = ?, v13 = ?
v12x2x1 = 250 m0 = 250 m = 1.61 m/s
t2t1155 s0155 s
v23x3x2 = 320 m = 1.18 m/s
t3t2272 s
v13x3x1
t3t1
x3x1 = (x2x1) + (x3x2)
x3x1 = 250 m + 320 m
x3x1 = 570 m
t3t1 = (t2t1) + (t3t2)
t3t1 = 155 s + 272 s
t3t1 = 427 s
v13x3x1 = 570 m = 1.33 m/s
t3t1427 s

#3

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.

Solution

72 km/h; −80 km/h; 0

How to arrive at the solution

x-axis with position x1 at the origin, position x2 positive, position x3 equal to x2, and position x4 at the origin.xO2x3x=1x4x=
t1 = 0, x1 = 0
t2 = 50 min, x2 = 60 km
t3t2 = 30 min, x3 = 60 km
t4t3 = 45 min, x4 = 0
v12 = ?, v34 = ?, v14 = ?
v12x2x1 = 60 km0 = 60 km
t2t150 min050 min
50 min = 50 (1/60 h) = 0.83 h
v1260 km = 72 km/h
0.83 h
v34x4x3 = 060 km = −60 km
t4t345 min45 min
45 min = 45 (1/60 h) = 0.75 h
v34−60 km = −80 km/h
0.75 h
v14x4x1 = 00 = 0 = 0
t4t1t4t1t4t1

#4

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?

Solution

8.19 m/s; 29.5 km/h

How to arrive at the solution

x-axis with position x1 at the origin, position x2 positive, and position x3 greater than x2.xO1x2x3x
t1 = 0, x1 = 0
t2 = 2.15 h, x2 = unknown, v12 = 8.83 m/s
t3 = unknown, x3x2 = 43.7 km, v23 = 7.35 m/s
v13 = ?
v13x3x1
t3t1
x3x1 = (x2x1) + (x3x2)
x3x2 = 43.7 km
t2t1 = 2.15 h0 = 2.15 h
v12 = 8.83 m/s
v12x2x1
t2t1
x2x1 = v12 (t2t1)
x2x1 = (8.83 m/s) (2.15 h)
x2x1 = (8.83 × 3.6 km/h) (2.15 h)
x2x1 = 68.3 km
x3x1 = (x2x1) + (x3x2)
x3x1 = 68.3 km + 43.7 km
x3x1 = 112.0 km
t3t1 = (t2t1) + (t3t2)
t2t1 = 2.15 h
x3x2 = 43.7 km
v23 = 7.35 m/s
v23x3x2
t3t2
(t3t2) v23 = x3x2
t3t2x3x2
v23
t3t243.7 km
7.35 m/s
t3t243.7 km
7.35 × 3.6 km/h
t3t2 = 1.65 h
t3t1 = (t2t1) + (t3t2)
t3t1 = 2.15 h + 1.65 h
t3t1 = 3.80 h
v13x3x1
t3t1
v13112.0 km = 29.5 km/h
3.80 h
v1329.5 m/s = 8.19 m/s
3.6

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