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In this Physics tutorial, you will learn:

- The steps to calculate the average speed in two and three dimensions?
- How can we calculate the instantaneous speed in two and three dimensions?
- The same for the average and instantaneous velocity in two and three dimensions

Tutorial ID | Title | Tutorial | Video Tutorial | Revision Notes | Revision Questions | |
---|---|---|---|---|---|---|

3.6 | Speed and Velocity in 2 and 3 Dimensions |

In the previous tutorial "Speed and Velocity in One Dimension," we explained the basics of these two concepts outlining at the same time the differences between them. Now, we will deal with **Speed** and **Velocity** in two and three dimensions. Obviously, the physical principles in both cases are the same; only the mathematical apparatus used to find the values is more complicated (just like when dealing with Displacement and Distance in two and three dimensions compared to the calculation of the same quantities in one dimension).

Therefore, in this tutorial we will use more a Mathematical-based approach than a Physical one because the physical principles of the quantities involved have already been explained in the previous tutorials of this section.

Let's start with the average speed first. Thus, basically, the calculation of the average speed in two dimensions consists on the following steps:

- Splitting the distance in small regular segments (exactly as we did when we were discussing about the calculation of distance in two dimensions).
- Finding the horizontal and vertical components for each segment. The values are all positive regardless the direction as we are dealing with distance, not displacement.
- Using the Pythagorean Theorem for calculating the distance in each segment (it represents the hypotenuse of each small right triangle formed).
- Finding the sum of all hypotenuses obtained at (3). This gives the total distance travelled.
- Dividing the distance found at (4) to the total time taken during the entire process.

The result obtained at (5) represents the average speed of a moving object in two dimensions.

Calculate the average speed of an object is it moves for 53 s according the path shown in the figure below.

We will use the procedure described above to find the average speed.

For simplicity, we divide the path into 4 segments and we appoint letters to each bordering point. They are AB, BC, CD and DE. Look at the figure below:

Let's calculate the components now. Thus, for the horizontal part of motion, we have

AB_{x} = |xB - xA| = |2m - 4m| = |-2m| = 2m

BC_{x} = |xC - xB| = |9m - 2m| = |7m| = 7m

CD_{x} = |xD - xC| = |13m - 9m| = |4m| = 4m

DE_{x} = |xE - xD| = |6m - 13m| = |-7m| = 7m

BC

CD

DE

The same procedure is used for the vertical direction as well:

AB_{y} = |yB - yA| = |8m - 3m| = |5m| = 5m

BC_{y} = |yC - yB| = |5m - 8m| = |-3m| = 3m

CD_{y} = |yD - yC| = |10m - 5m| = |5m| = 5m

DE_{y} = |yE - yD| = |11m - 10m| = |1m| = 1m

BC

CD

DE

Now, we will calculate the distance travelled in each segment:

AB = √**AB**^{2}_{x}+AB^{2}_{y} = √**(2m)**^{2}+(5m)^{2} = √**4m**^{2}+25m^{2} = √**29m**^{2} = 5.4m

BC = √**BC**^{2}_{x}+BD^{2}_{y} = √**(7m)**^{2}+(3m)^{2} = √**49m〗**^{2}+9m^{2} = √**58m**^{2} = 7.6m

CD = √**CD**^{2}_{x}+CD^{2}_{y} = √**(4m)**^{2}+(5m)^{2} = √**16m**^{2}+25m^{2} = √**41m**^{2} = 6.4m

DE = √**DE**^{2}_{x}+DE^{2}_{y} = √**(7m)**^{2}+(1m)^{2} = √**49m**^{2}+1m^{2} = √**50m**^{2} = 7.1m

BC = √

CD = √

DE = √

Now, let's calculate the total distance. We have:

s_{total} = AB + BC + CD + DE

= 5.4m + 7.6m + 6.4m + 7.1m

= 26.5m

= 5.4m + 7.6m + 6.4m + 7.1m

= 26.5m

The average speed therefore is

< v > = *s*_{total}*/**t*_{total} = *26.5m**/**53s* = 0.5 *m**/**s*

For **instantaneous speed**, we use again the same procedure as in the one-dimensional motion, i.e. after calculating the distance in a small interval around the required point (using the Pythagorean Theorem for the two given directions), we divide it by the given time interval. The result obtained gives the instantaneous speed in two dimensions.

The procedure for finding the (average) velocity in two dimensions is much shorter than the one used for calculating the average speed. We need only the coordinates of the initial and final position; none of the in-between coordinates are needed.

1. First we find the x and y-coordinates of the starting and ending point (x_{i}, y_{i}, x_{f}, and y_{f})

2. Then we find the displacement for each direction by using the equations

∆x*⃗* = x*⃗*_{f} - x*⃗*_{i}

and

∆y*⃗* = y*⃗*_{f} - y*⃗*_{i}

3. Afterwards we find the magnitude of the total displacement ∆r*⃗* by using the equation

|∆r*⃗*| = √**∆x***⃗*^{2} + ∆y*⃗*^{2}

4. Finally, we calculate the magnitude of the average velocity using the equation

|< v*⃗*>| = *|∆r**⃗*|*/**t*

where t is the time taken for the entire motion.

We can illustrate this point by considering again the numerical example in the previous paragraph.

1. We have xA = 4m, xE = 6m, yA = 3m and yE = 11m.

2. Hence,

∆x*⃗* = x*⃗*_{f} - x*⃗*_{i}

= x*⃗*_{E} - x*⃗*_{A}

= 6m - 4m

= 2m

= x

= 6m - 4m

= 2m

and

∆y*⃗* = y*⃗*_{f} - y*⃗*_{i}

= y*⃗*_{E} - y*⃗*_{A}

= 11m - 3m

= 8m

= y

= 11m - 3m

= 8m

3. The total displacement therefore is

|∆r*⃗* | = √**∆x***⃗*^{2} + ∆y*⃗*^{2}

= √**(2m)**^{2} + (8m)^{2}

= √**4m**^{2} + 64m^{2}

= √**68m**^{2}

= 8.25 m

= √

= √

= √

= 8.25 m

4. Hence, the total average velocity is

|< v*⃗* >| = *|∆r**⃗*|*/**t*

=*8.25m**/**26.5s*

= 0.31*m**/**s*

=

= 0.31

With **Instantaneous Velocity**, the procedure is the same as in the case of instantaneous speed. The only difference is the sign of result, which unlike in the instantaneous speed, here can be positive or negative depending on the sign of the displacement.

Everything discussed for Speed and Velocity in two dimensions is also true for these quantities in three dimensions. We have only to add a new dimension (coordinate) z in the calculations. Let's see an example.

Calculate the average speed and average velocity for the airplane shown in the figure if both trajectories AB and BC are half circles. The radius of AB is 350 m and that of BC is 560 m. The flight AB is done according the horizontal plane xOy and the flight BC according the vertical plane yOz. Take π = 22/7. The total time of flight is 10 s.

First, let's start with the average speed. We write as R_{1} the radius of the arc AB and R2 that of the arc BC. Thus, for the arc AB we have

s_{AB} = *1**/**2* Circumference of the circle by diameter AB

=*1**/**2* × 2πR_{1}

= πR

=*22**/**7* × 350 m

= 1100 m

=

= πR

=

= 1100 m

For the arc BC we have:

s_{BC} = *1**/**2* Circumference of the circle by diameter BC

=*1**/**2* × 2πR_{2}

= πR_{2}

=*22**/**7* × 560 m

= 1760 m

=

= πR

=

= 1760 m

Therefore, the total distance travelled by the airplane, is:

s_{tot} = s_{AB}+s_{BC}

= 1100 m+1760 m

= 2860 m

= 1100 m+1760 m

= 2860 m

Hence, the average speed of the airplane during the entire motion is

v ≥ *s*_{total}*/**t*_{total}

=*2860m**/**10s*

= 286*m**/**s*

=

= 286

As for the displacement, we can see from the figure that we have two displacements combined together. The first is the displacement AB*⃗*, which lies according the Oy-direction only, although the arc AB lies in the plane xOy. This is because we are interested only in the diameter AB of the first half-circle, which represents the shortest path (the displacement) from A to B.

The same thing can be said for the displacement BC as well, only that the displacement vector BC*⃗* lies in the Oz-direction, despite the fact that the arc BC lies in the yOz plane.

Hence, combining the findings above, it is obvious the total displacement vector AC*⃗* lies in the yOz plane and it represents the hypotenuse AC of the right triangle ABC where AB and BC are its legs.

Mathematically, we have:

|∆r*⃗*_{1}| = |AB*⃗*|

= 2 × r_{1}

= 2 × 350 m

= 700 m

= 2 × r

= 2 × 350 m

= 700 m

and

|∆r*⃗*_{2}| = |BC*⃗*|

= 2 × r_{2}

= 2 × 560 m

= 1120 m

= 2 × r

= 2 × 560 m

= 1120 m

Therefore, the magnitude of the total displacement |∆r*⃗*| = |AC*⃗*| is

|∆r*⃗*_{tot} | = √**∆r***⃗*_{1}^{2}+∆r*⃗*_{2}^{2}

= √**(700 m)**^{2}+(1120 m)^{2}

= √**490,000 m**^{2}+1,254,000 m^{2}

= √**1,744,400 m**^{2}

≈1321 m

= √

= √

= √

≈1321 m

Hence, the average velocity of the airplane during the entire motion is

v*⃗* ≥ *|∆r**⃗*_{tot}|*/**t*_{tot}

=*1321m**/**10s*

= 132.1*m**/**s*

≈ 132*m**/**s*

=

= 132.1

≈ 132

(The result is rounded up to the nearest whole number in order to write it with the correct number of significant figures (see the Physics tutorial Significant Figures and Their Importance).

**Remark!** In all situations discussed so far, we have assumed the motion to be either uniform (i.e. at the same speed) or it was "modulated" in such a way that the average values of distance, displacement, speed and velocity replaced the actual values. In this way, the calculations became easier and shorter.

Enjoy the "Speed and Velocity in 2 and 3 Dimensions" physics tutorial? People who liked the "Speed and Velocity in 2 and 3 Dimensions" tutorial found the following resources useful:

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- Continuing learning kinematics - read our next physics tutorial: The Meaning of Acceleration. Constant and Non-Constant Acceleration. Gravitational Acceleration

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