__Speed, velocity and acceleration__

As you know, ** distance** and

**have distinctly different meanings. The same can be said about**

*displacement***and**

*speed***.**

*velocity***(**

*Speed***v**) is a

**scalar quantity**that shows "how fast an object is moving."

__Speed____can be thought of as the rate at which an object covers a given__. A fast-moving object has a higher

*distance***and covers a**

*speed***in the same amount of**

*greater distance***compared to a slow-moving object. An object which is**

*time***not moving**has a

**zero speed**. Therefore,

\(v=\frac{d}{t}\)

where **d** is the distance (in meters) and **t** is the time (in seconds)

From here you can draw the unit of speed. Since the **distance** is measured in **metres** and **time** is measured in **seconds**, the unit of **speed** is **meters per second** or in short, **m/s.**

For long distances measured in **kilometres** **(km)**, it would be more appropriate to calculate the **time** in **hours (h)** instead of **seconds (s)**. Therefore, another unit for **speed (km/h)** is obtained. The conversion between these two units is as follows:

\(3.6\; {km \over h} \times {1000\;m \over 1\;km} \times {1\;h\over3600\;s}= 1\; m/s\)

It is important to note that the abovementioned equation of **speed** is valid in these 2 situations:

1. When the object is moving by the **same (unchanging) speed** during the whole trip. This type of motion is called **uniform motion** and this unchanging speed is called **constant speed.**

2. When the motion is so irregular that we are not interested to calculate the **speed at every instant** (this is called **instantaneous speed**) but instead, we prefer to calculate the **average speed** (exactly in the same way you calculate all the other average quantities in science), by using the equation:

\(v(average)=\frac{d(total)}{t(total)}\)

The meaning of the **average speed** concept is as follows: "If instead of the real irregular motion you did during a trip you had moved at **constant speed** by magnitude as the one of the **average speed** calculated, you would have covered **the same distance** at **the same time** as you did during your irregular motion.

** Example 1:** A girl goes from home at school for 15 minutes. The distance from her house to the school is 1.8km. What is her average speed: a) in m/s; b) in km/h.

**Solution:** The information regarding her type of motion in missing. Therefore, we consider the girl's motion as uniform by using as a speed parameter the value of her average speed.

a) First convert 1.8 km to 1800 m by multiplying it by 1000 (1km=1000m) and 15 minutes to 900 s by multiplying it by 60 (1 min=60 s). Thus,

\(v(average)=\frac{d(total)}{t(total)}=\frac{1800\;m}{900\;s}=2\;m/s\)

b) Convert 15 min to 0.25 h by dividing it by 60 (1 min = 1/60 h). Thus,

\(v(average)=\frac{d(total)}{t(total)}=\frac{1.8\;km}{0.25\;h}=7.2k\;m/h\)

On the other hand, **velocity** \((\vec v)\) is a **vector quantity** that represents "* the rate at which an object changes its position*." Imagine a person moving rapidly - one step forward and one step back - always returning to his/her original

**starting position**. While this might result in a frenzy of activity, it would result in a

**zero velocity**. Because the person always returns to the

**original position**, the motion would never result in a

**change in position**. Since

**velocity**is defined as

*this motion results in*

**the rate at which the position changes,****zero velocity**. If a person in motion wishes to maximize his/her

**velocity,**then he/she must make every effort to maximize the amount that he/she is

**displaced**from the

**original position**. Every step must go into moving that person further from where he or she started. To have a

*on this, the person should*

**maximal effect****never change directions**or

**turning back**.

**Velocity is a vector quantity**. As such, **velocity** is * direction aware*. When evaluating the

**velocity**of an object, one must keep track of

**direction**. It would not be enough to say that an object has

**a velocity of 30km/h**. One must include direction information in order to fully describe the velocity of the object. For instance, you must describe an object's velocity as being

**30 km/h, west**. This is one of the essential differences between

**speed**and

**velocity**.

__Speed is a scalar quantity and does not keep track of direction; velocity is a vector quantity and is direction aware.__To calculate the **velocity,** the following equation is used:

\(\vec{v}=\frac{\vec{\Delta{x}}}{t}\)

where \((\vec v)\) is the **velocity** of object, \((\Delta \vec x)\) is the **change in position (displacement)** and (**t)** is the **time of motion**. From here, we can obtain the **unit of velocity** which again is **m/s (**or **km/h)** as that of the **speed**, because **displacement** is measured in **metres** (or in **kilometres** for **long distances**) and **time** is measured in **seconds** (or in **hours** for **long time**). To make clear the difference between **displacement** and **distance** and also between **speed** and **velocity,** look at the following example:

** Example 2:** A boy moves 40 m east and then 30 m west at the same rate during 10 seconds. Calculate:

a) the distance (d) covered by the boy,

b) his displacement \((\Delta \vec x)\),

c) the speed \((v)\) of the boy,

d) his velocity \((\vec v)\)

__Solution:__

a) As the boy has moved 40 m east and 30 m west, his total motion (distance covered) is:

\(d=40\;m+30\;m=70\;m\)

b) The boy has initially moved 40 m due east. Then he has turned back (due west) by 30 m. Taking the east direction as positive and west direction as negative, we can obtain the total displacement

\(\vec{\Delta{x}_{total}}=\vec{\Delta{x}_{east}}+\vec{\Delta{x}_{west}=40\;m+(-30\;m)=10\;m}\)

As the result is positive (we got the east direction as positive) the correct answer for the total displacement would be: **10 m due east. **

c) Speed \((v)\) can be easily calculated through the equation

\(v=\frac{d}{t}\)

Thus, since the time \((t)\) is 10s, we obtain for the speed:

\(v=\frac{d}{t}=\frac{70\;m}{10\;s}=7\;m/s\)

d) Likewise, velocity \((\vec v)\) is calculated by using the equation

\(\vec{v}=\frac{\vec{\Delta{x}}}{t}=\frac{10\;m}{10\;s}=1\;m/s\space(due\space{east})\)

Yet, the concept of **average velocity** is introduced at the same way as for the **average speed** (because it is hard to keep always the same rate of motion). Thus, the **average velocity** is calculated by using the equation:

\(average\space{velocity}\space(\vec{v}_{average})=\frac{total\space{displacement(\Delta{\vec{x}})}}{total\space{time(\Delta{t})}}\)

__The meaning of acceleration__

Let's take a simple example to describe a particular situation. An object is moving at 1 m/s at a certain instant to a certain direction. Then it suddenly starts to **increase its velocity uniforml**y. Thus, after the first second its **velocity** becomes 3 m/s, after another second its **velocity** becomes 5 m/s, again after another second its **velocity** becomes 7 m/s and so on. As we see, the rate of **object's increase in velocity** is the same: * 2 m/s every second* (7-5 =5-3 =3-1=2). Therefore, we say that the object has experienced a

**constant acceleration**by 2 m/s every second to the direction of motion or simply,

**its velocity has increased by 2 m/s**^{2}. From here, we can get the meaning of **acceleration**. **Acceleration, \((\vec a)\)** is ** the rate of velocity change** or otherwise,

**. Thus,**

__acceleration is the change in velocity in the unit time__\(\vec{a}=\frac{\vec{\Delta{v}}}{t}=\frac{\vec{v_{final}}-\vec{v_{initial}}}{t}\)

The arrow above the symbol \((a)\) shows that **acceleration is a vector quantity**. As you may know, when a vector quantity (**change in velocity**) is divided to a **scalar** one (time), the result is a **vector again**. As such, we have to know the **direction of velocity change** in order to get a full meaning of the **acceleration**.

\(\vec{a}=\frac{\vec{\Delta{v}}}{t}=\frac{\vec{v_{final}}-\vec{v_{initial}}}{t}=\frac{\vec{v}-\vec{v_0}}{t}\)

where \(\vec v\) and \(\vec v_0\) usually refer to **final** and **initial velocity** respectively.

The **velocity** may not always change **at the same rate**. This type of motion is called a **non-uniform accelerated motion**. But here we will consider only situations where the **velocity changes at the same rate**. Such a motion is called **uniformly accelerated motion**. This means that in the entire situation studied (example or exercise whatever it be), the **acceleration** will not change.

The opposite of **acceleration** is called **deceleration**. it means a slowing down motion or a **negative acceleration**. In such a case, the **magnitude of velocity decreases every second** even though the object may be still moving at the same direction as before. Therefore, **negative acceleration** does not mean that the object is changing its direction of motion but instead, it is **reducing the magnitude of its velocity**.

From here, we can see that * if the velocity is not changing, the magnitude of acceleration is zero*. Therefore, it can be said that uniform motion is a special case of non-uniform accelerated (decelerated) motion.

** Example 3:** A car is moving at 15 m/s. Suddenly, the driver sees an obstacle in front of him and applies the brakes. Calculate the time needed to stop if the average deceleration caused by the friction between the wheels and asphalt is 6 m/s

^{2}.

__ Solution:__ As mentioned before, the term deceleration refers to a negative acceleration (as the car slows down). Instead of writing deceleration = 6 m/s

^{2}we can write acceleration = -6 m/s

^{2}. Furthermore, we have an extra (hidden) information here. In the exercise,

**the time needed to stop**is required. Therefore, the

**final velocity**will be zero. Hence,

\(\vec{a}=\frac{\vec{\Delta{v}}}{t}=\frac{\vec{v_{final}}-\vec{v_{initial}}}{t}\implies{t}=\frac{\vec{v_{final}}-\vec{v_{initial}}}{\vec{a}}=\frac{0\;m/s-15\;m/s}{-6\;m/s}=\frac{-15\;m/s}{-6\;s}=2.5\;s\)

**Note:** In the above example we have supposed that the motion is **linear**. When the motion is **non-linear** there is a more complicated situation that is not studied in this topic.