## Table of Contents

To understand the difference between speed and velocity, you will need to grasp the difference between distance and displacement, which can be found here.

## Speed

**Speed** is the distance moved per unit time.

$$\text{speed} = \frac{\text{distance}}{\text{time}}$$

or:

$$\text{speed} \, = \, \frac{d}{t}$$

, where d is distance travelled and $t$ is time taken

- SI unit is metre per second ($m \, s^{-1}$)
- Scalar quantity

### Average Speed

**Average speed**, $\left< \text{speed} \right>$ can be calculated using:

$$ \begin{aligned} \left< \text{speed} \right> &= \frac{\text{total distance travelled}}{\text{total time taken}} \\ &= \: \frac{\Delta x}{\Delta t} \end{aligned}$$

**Note**: The upper case symbol for the Greek letter delta, $\Delta$ is used to mean a change in a quantity.

- When a car covers a distance of 300 km in a duration of five hours, its average speed can be calculated as $\frac{300 \text{ km}}{5 \text{ h}}$, resulting in 60 km/h. It’s important to note that the speedometer won’t consistently display 60 km/h throughout the entire journey and may fluctuate significantly from this value. This is why we refer to the average speed. If, hypothetically, a car were to maintain a constant speed of 60 km/h for the entire five-hour period, the total distance covered would still be 300 km.

### Instantaneous Speed

**Instantaneous speed** is the speed at any instant

- Measured by a speedometer

$$\text{Rate of change of distance with respect to time at that instant} = \frac{dx}{dt}$$

### Uniform Speed

When there is a constant rate of change in the object’s distance covered per unit of time, we say that the object is experiencing constant or uniform speed.

Time (s) | Distance (m) | Change in Distance (m) |
---|---|---|

0 | 0 | 0 |

1 | 10 | $10 – 0 = 10$ |

2 | 20 | $20 – 10 = 10$ |

3 | 30 | $30 – 20 = 10$ |

## Velocity

**Velocity** (v) of an object is the rate of change of displacement with respect to time.

$$\text{velocity} = \frac{\text{displacement}}{\text{time}}$$

or:

$$\text{v} \, = \, \frac{s}{t}$$

,where $s$ is displacement and $t$ is time taken

- SI unit is metre per second ($m \, s^{-1}$)
- Vector quantity – As velocity is a vector quantity, you have to specify its magnitude and direction to completely describe it.
- Magnitude of velocity at a given point is given by instantaneous speed at that point.
- Direction of velocity is tangential to path of object.
**Positive Velocity:**Indicates motion in the positive direction (e.g., forward or right).**Negative Velocity:**Indicates motion in the negative direction (e.g., backward or left).

Displacement is the distance traveled in a given direction, treated as a vector, while distance is a scalar quantity. The significance of direction becomes apparent when considering speed. For example, if two trains move north at a speed of 20 m/s, they share both the same speed and velocity (20 m/s in the northward direction). However, if one travels north and the other south, even with equal speeds, their velocities differ due to their distinct motion directions.

### Average Velocity

$\left< v \right>$ can be calculated using:

$$\left< v \right> = \frac{\text{total displacement}}{\text{total time taken}}$$

$$ \begin{aligned} \left< v \right> &= \frac{\text{total distance travelled}}{\text{total time taken}} \\ &= \: \frac{\Delta s}{\Delta t} \end{aligned}$$

### Instantaneous Velocity

Instantaneous velocity is given by the following formula:

$$\text{Rate of change of displacement with respect to time at that instant} = \frac{ds}{dt}$$

### Relationship Between Velocity & Direction of Motion:

**Positive Velocity:** Indicates motion in the chosen positive direction.

- Example: A car moving forward with positive velocity.

**Negative Velocity:** Indicates motion in the opposite (negative) direction.

- Example: A car moving backward with negative velocity.

## Worked Examples

### Example 1

A boy ran a distance of 100 metres in 11.32 seconds. What was his average speed?

**Click here to show/hide answer**

From above, we have:

$$\begin{aligned} \left< \text{speed} \right> &= \frac{\text{total distance travelled}}{\text{total time taken}} \\ &= \frac{100}{11.32} \\ &= 8.83 \text{ m s}^{-1} \end{aligned}$$

### Example 2

The speed of an object is considered uniform or constant when it travels at a consistent rate in a straight line. However, if the object follows a curved path, its velocity is not uniform. Why is this the case?

**Click here to show/hide answer**

The reason lies in the change of direction. In a curved path, the object is constantly changing its direction, even if its speed remains the same. Velocity is a vector quantity, combining speed and direction. Therefore, if the direction changes, even with a constant speed, the velocity is not uniform. This is a result of the vector nature of velocity and the requirement to consider both speed and direction to fully describe the motion of an object.

### Example 3

A car travels 5 km due east and makes a U-turn back to travel a further distance of 3 km. The car completes the journey in 0.3 hours.

Find:

- the distance covered;
- the displacement of the car;
- the average speed;
- the average velocity.

**Click here to show/hide answer**

(a) The distance covered is $5 + 3 = 8 \text{ km}$.

(b) The displacement is $5 -3 = 2 \text{ km due east of the starting point}$.

(c)

$$\begin{aligned} \left< \text{speed} \right> &= \frac{\text{total distance travelled}}{\text{total time taken}} \\ &= \frac{8}{0.3} \\ &= 26.7 \text{ km h}^{-1} \end{aligned}$$

(d)

$$\begin{aligned} \left< v \right> &= \frac{\text{total displacement}}{\text{total time taken}} \\ &= \frac{2}{0.3} \\ &= 6.67 \text{ km h}^{-1} \text{ due east of starting point} \end{aligned}$$

### Example 4

What is the difference between speed and velocity?

**Click here to show/hide answer**

Speed is a scalar while velocity is a vector.

Speed is a rate of change of distance while velocity is the rate of change of displacement.

### Example 5

For an object that is moving at a constant velocity, is it necessary for it to be moving in a straight line?

**Click here to show/hide answer**

Velocity is a vector quantity – both magnitude and direction are accounted for. For an object to have constant velocity, it is necessary that both the magnitude and the direction of the velocity vector be kept constant. If either magnitude or direction is changing, the velocity will not remain constant.

Hence, for an object to have a constant velocity, it is necessary for it to be moving in a straight line at a constant speed.

On the other hand, speed can remain constant if direction is changed and magnitude is kept constant, as speed is a scalar quantity.

### Example 6

The average velocity of a particle moving in one dimension has a positive value. Is it possible for the instantaneous velocity to have been negative at any time in the interval? Suppose the particle started at the origin x = 0. If its average velocity is positive, could the particle ever have been in the – x region of the axis?

**Click here to show/hide answer**

Yes. Yes, if the particle winds up in the +x region at the end.

### Example 7

Can the instantaneous velocity of an object at an instant of time ever be greater in magnitude than the average velocity over a time interval containing the instant? Can it ever be less?

**Click here to show/hide answer**

Yes.

Yes.

### Example 8

If an object’s average velocity is nonzero over some time interval, does that mean that its instantaneous velocity is never zero during the interval? Explain your answer.

**Click here to show/hide answer**

No. Consider a sprinter running a straight-line race. His average velocity would simply be the length of the race divided by the time it took for him to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero.

### Example 9

A particle moves according to the equation x = 10 t^{2}, where x is in meters and t is in seconds. (a) Find the average velocity for the time interval from 2.00 s to 3.00 s. (b) Find the average velocity for the time interval from 2.00 s to 2.10 s.

**Click here to show/hide answer**

(a) The average velocity $v_{\text{avg}}$ for the time interval from $t_1$ to $t_2$ is given by the formula:

$$v_{\text{avg}} = \frac{\Delta x}{\Delta t}$$

where $\Delta x$ is the change in position and $\Delta t$ is the change in time.

Given the equation $x = 10t^2$, the position at time $t$ is $x = 10t^2$. Now, let’s find the change in position $\Delta x$ and the change in time $\Delta t$ for the interval from $t = 2.00 \text{ s}$ to $t = 3.00 \text{ s}$.

$$\begin{aligned} \Delta x &= x_2 – x_1 \\ &= (10t_2^2) – (10t_1^2) \end{aligned}$$

$$\begin{aligned} \Delta t &= t_2 – t_1 \\ &= 3.00 \, \text{s} – 2.00 \, \text{s} \end{aligned}$$

Now, let’s substitute these values into the formula for average velocity:

$$\begin{aligned} v_{\text{avg}} &= \frac{(10t_2^2) – (10t_1^2)}{t_2 – t_1} \\ &= \frac{10(3)^2 – 10(2)^2}{3 – 2} \\ &= 50 \text{m s}^{-1} \end{aligned}$$

(b) We need to find the average velocity for the time interval from $t = 2.00 \text{ s}$ to $t = 2.10 \text{ s}$. Using the same formula as before, with the appropriate values – $t_1 = 2.00 \text{ s}$ and $t_2 = 2.10 \text{ s}$:

$$\begin{aligned} v_{\text{avg}} &= \frac{(10t_2^2) – (10t_1^2)}{t_2 – t_1} \\ &= \frac{10(2.1)^2 – 10(2)^2}{2.1 – 2} \\ &= 41 \text{ m s}^{-1} \end{aligned}$$

### Example 10

A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s.

(a) What is her average speed over the entire trip?

(b) What is her average velocity over the entire trip?

**Click here to show/hide answer**

(a)

Average speed $v_{avg}$ is defined as the total distance traveled divided by the total time taken.

The total distance traveled is the sum of the distances from A to B and from B to A. Let the distance from A to B be $x$

$$\text{Total distance} = 2x$$

The total time taken is the sum of the time taken to travel from A to B, $ t_{AB}$, and the time taken to travel from B to A, $t_{BA}$.

$$t_{AB} = \frac{d_{AB}}{v_{AB}} = \frac{x}{5}$$

$$t_{BA} = \frac{d_{BA}}{v_{BA}} = \frac{x}{3}$$

Sub into:

$$\begin{aligned} v_{avg} &= \frac{d_{AB} + d_{BA}}{t_{AB} + t_{BA}} \\ &= \frac{2x}{\frac{x}{5} + \frac{x}{3}} \\ &= \frac{2}{\frac{8}{15}} \\ &= 3.75 \text{ m s}^{-1} \end{aligned}$$

(b)

Average velocity $v_{avg}$ is defined as the total displacement traveled divided by the total time taken.

Total displacement is ZERO – she went back to her starting position.

Hence, average velocity is $0 \text{ m s}^{-1}$

### Example 11

A hare and a tortoise compete in a race over a course 1.00 km long. The tortoise crawls straight and steadily at its maximum speed of 0.200 m/s towards the finish line. The hare runs at its maximum speed of 8.00 m/s toward goal for 0.800 km and then stops to taunt the tortoise. How close to the goal can the hare let the tortoise approach before resuming the race, which the tortoise wins in a photo finish? Assume that, when moving, both animals move steadily at their respective maximum speeds.

**Click here to show/hide answer**

Lets find the time required for the hare to run the remaining 0.2 km.

$$\begin{aligned} t &= \frac{d}{v} \\ &= \frac{200}{8} \\ &= 25 \text{ s} \end{aligned}$$

Lets find the distance that the tortoise can travel in 25 seconds:

$$\begin{aligned} d &= v \times t \\ &= 0.2 \times 25 \\ &= 5 \text{ m} \end{aligned}$$