# Speed, Velocity and Acceleration

In linear or translational motion a whole body moves from one point to another which is different from rotational or angular motion where a whole body rotates about a fixed axis.

When simply talking about speed, velocity and acceleration, we are talking about linear speed, linear velocity and linear acceleration respectively.

We usually don't use the term linear or translational for position, speed, velocity and acceleration but while comparing with or using with rotational or angular motion we use linear or translational to distinguish with angular motion.

In straight line motion, that is one dimensional motion such as along x-axis, a single number could determine the position of a particle but in two dimensional motion such as in xy-plane the position is determined by position vector drawn from origin of coordinate system to the location of the particle. Therefore, vectors are extensively used in two or three dimensional motion and very good understanding of vectors is required.

In two or three dimensional motion the quantities such as displacement, velocity and acceleration no longer lie along straight line.

## Linear Speed

The linear speed of a body is the total distance traveled by the body per unit time which doesn't matter whether the path of motion is a curve or a straight line.

We do not take any account of direction of motion in speed. The speed is also the magnitude of instantaneous velocity at any particular point of the path of motion.

To find the distance traveled per unit time you divide the total distance traveled $s$ by the total time taken $t$ to cover that distance. The speed is always a positive quantity.

${\rm{speed}} = \left| {\frac{s}{t}} \right|$

You may think the speed and velocity are the same things in everyday life but this is not the case in Physics as there are physical quantities with or without direction.

## Linear Velocity

The velocity is the rate of change of displacement. The displacement is different from distance and always has a direction and therefore a velocity is always associated with a direction.

In speed a body can move in multiple directions (it means it can turn in various directions during the motion) but in velocity the particle should move without turning or changing the direction.

So, velocity is a vector quantity and speed is a scalar quantity. Note that speed is always positive. If the direction of the particle changes not the magnitude of velocity, we still say that the velocity changes as you should always think both magnitude and direction for velocity.

In Figure 1 a car is moving in a straight line from point ${{p}_{1}}$ to point $p_2$ along x-axis. The distance between the points is $\Delta s={{s}_{2}}-{{s}_{1}}$. Let the time at point $p_1$ is $t_1$ and at point $p_2$ is $t_2$, so the total time during the motion between the points is $\Delta t={{t}_{2}}-{{t}_{1}}$. Now the magnitude of velocity is the total distance travelled in a particular direction divided by the total time taken during the displacement, which is,

$v=\frac{{{s}_{2}}-{{s}_{1}}}{{{t}_{2}}-{{t}_{1}}}=\frac{\Delta s}{\Delta t} \tag{1} \label{1}$

Eq. \eqref{1} gives the average velocity between the points ${{p}_{1}}$ and ${{p}_{2}}$. When we simply say the term "velocity" we generally refer to the instantaneous velocity. Instantaneous velocity is the velocity at any instant of time during the motion.

For example, if you say the velocity of a car is $13\text{m/s}$ at a particular point of time, that velocity is instantaneous velocity. To find instantaneous velocity, let $\Delta t$ approaches zero in Eq. \eqref{1}. When $\Delta t$ approaches zero, the point $p_2$ moves closer and closer to the point $p_1$. So the magnitude of the instantaneous velocity in the limit of $\Delta t$ approaches zero is,

$v=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta s}{\Delta t}=\frac{ds}{dt} \tag{2} \label{2}$

Note that $\Delta t$ approaches zero means the difference ${{t}_{2}}-{{t}_{1}}$ approaches zero which also makes the displacement, $\Delta s$ very small.

In Figure 2 the displacement-time graph is a straight line where $s$ is the displacement and $t$ is the time. If the velocity is constant the s-t graph is always a straight line with constant slope.

You should know that when the term "velocity" is used we mean "instantaneous velocity". When the velocity changes the s-t graph is always a curve as in Figure 3 and the tangent at each point on the curve represents the slope at that point. The slope gives the instantaneous velocity or simply velocity at any point on the curve.

### Velocity in Vector Form

In two or three dimensional situations the motion is not along a straight line. Consider an object which moves from a point ${{p}_{1}}$ to another point ${{p}_{2}}$ in two dimensional motion (xy-plane) as shown in Figure 4. In two or three dimensional motion the position is determined by the position vector.

Let the position vector of point ${{p}_{1}}$ be ${\vec{r}_{1}}$ and that of ${{p}_{2}}$ be ${\vec{r}_{2}}$. Now the net displacement in the direction of motion is , $\Delta \vec r = {{\vec{r}}_{2}}-{{\vec{r}}_{1}}$

Suppose the object's initial time at point ${{p}_{1}}$ is ${{t}_{1}}$ and at ${{p}_{2}}$ is ${{t}_{2}}$. So, the total time during the motion between the points is $\Delta t={{t}_{2}}-{{t}_{1}}$. The velocity is the total displacement divided by the time taken during the displacement. Therefore the average velocity in vector form is,

${{\vec{v}}_{av}}=\frac{{{\vec{r}}_{2}}-{{\vec{r}}_{1}}}{{{t}_{2}}={{t}_{1}}}=\frac{{\Delta \vec r}}{\Delta t} \tag{3} \label{3}$

The subscript indicates that the velocity is average velocity. When $\Delta t$ approaches zero the point ${{p}_{1}}$ moves closer and closer to the point ${{p}_{2}}$. Can you say the point ${{p}_{2}}$ moves closer and closer to the point ${{p}_{1}}$?

In this case $\Delta t$ approaches zero means that both ${{t}_{1}}$ and ${{t}_{2}}$ are almost equal, so we can also say that the point ${{p}_{2}}$ moves closer and closer to the point ${{p}_{1}}$ and vice versa. The instantaneous velocity in vector form in the limit that $\Delta t$ approaches zero is given as,

${{\vec{v}}_{\operatorname{ins}}}=\vec{v}=\frac{d\vec{r}}{dt}$

The subscript "ins" is for the instantaneous velocity but we generally ignore this subscript and think the velocity as instantaneous velocity without that subscript. We also call the instantaneous velocity in vector form velocity vector.

## Linear Acceleration

What if the velocity changes with time? Then we have acceleration. Acceleration is the rate of change of velocity similar to the rate of change of displacement is velocity.

In Figure 5 a car has velocity ${{v}_{1}}$ at point ${{p}_{1}}$ and ${{v}_{2}}$ at point ${{p}_{2}}$ along x-axis. The earlier velocity ${{v}_{1}}$ is the initial velocity ${{v}_{i}}$ and the final one is ${{v}_{2}}$ which is ${{v}_{f}}$. The subscripts $i$ and $f$ represent the initial and final velocities respectively.

The change in velocity is the final velocity minus the initial velocity, that is, $\Delta v={{v}_{2}}-{{v}_{1}}={{v}_{f}}-{{v}_{i}}$. The time of the car at point $p_1$ is $t_1$ and at point $p_2$ is $t_2$. The time taken by the car to move from point $p_1$ to the point $p_2$ is $\Delta t={{t}_{2}}-{{t}_{1}}$. We express the rate of change in velocity as,

${{a}_{av}}=\frac{{{v}_{f}}-{{v}_{i}}}{\Delta t}=\frac{\Delta v}{\Delta t} \tag{4} \label{4}$

Notice that the subscript "$av$" represents the average acceleration. The acceleration in Eq. \eqref{4} is average acceleration because we don't know the rate of change of velocity is constant or not throughout the motion.

The whole process going here is, suppose the velocity at point ${{p}_{1}}$ is $5\text{m/s}$ and after $1\text{s}$ it is $10\text{m/s}$, and again after 1s it is $12\text{m/s}$. The acceleration during the first $1\text{s}$ is $\frac{(10-5)\text{m/s}}{1\text{s}}=5\text{m/}{{\text{s}}^{\text{2}}}$ and during the next $1\text{s}$ is $\frac{(12-10)\text{m/s}}{1\text{s}}=2\text{m/}{{\text{s}}^{\text{2}}}$, so the average of these two accelerations is $\frac{(5+2)\text{m/s}}{2}=3.5\text{m/}{{\text{s}}^{\text{2}}}$ which is the same as $\frac{(12-5)\text{m/s}}{2\text{s}}=3.5\text{m/}{{\text{s}}^{\text{2}}}$.

When $\Delta t$ approaches zero, the acceleration is instantaneous acceleration similar to instantaneous velocity. The instantaneous acceleration is the acceleration at a particular instant of time.

The instantaneous acceleration gives the acceleration at any point at any instant of time, so we mean the term instantaneous acceleration as a single word "acceleration". For example, if we say that a car has acceleration $3\text{m/}{{\text{s}}^{\text{2}}}$ at any point, that acceleration is instantaneous acceleration at that point.

The instantaneous acceleration is the average acceleration in Eq. \eqref{4} in the limit that $\Delta t$ approaches zero (the point ${{p}_{1}}$ is very close to the point ${{p}_{2}}$).

${{a}_{\text{ins}}}=a=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{{{v}_{f}}-{{v}_{i}}}{\Delta t}=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta v}{\Delta t}=\frac{dv}{dt} \tag{5} \label{5}$

The subscript $\text{ins}$ represents instantaneous for instantaneous acceleration which is not always necessary.

The graph of ${{v}_{x}}-t$ is always a straight line if the acceleration is constant indicating the constant slope. You may know that the slope at any point on the ${{v}_{x}}-t$ graph gives the instantaneous acceleration at that point. In Figure 6 the slope is constant and acceleration is the same (constant) throughout the motion.

What if the acceleration is not constant? You will find the result of ${{v}_{x}}-t$ graph to be some kind of curve and the slope at any point on the curve gives the acceleration at that point as in Figure 7

### Acceleration in Vector Form

In two or three dimensional motion the quantities displacement, velocity and acceleration are not along a straight line

Acceleration is a vector quantity. Velocity is also a vector quantity and if only the direction changes not the magnitude there is still acceleration. Consider that a car speeds up in a curved path and in this case both direction and magnitude of the velocity change, see in Figure 8. The magnitude of velocity at any point is the speed at that point.

The velocity in Figure 8 at point ${{p}_1}$ is $\vec{v}_1$ and at point ${{p}_2}$ is $\vec{v}_2$. The change in velocity is $\vec {\Delta v} = {\vec v _2} + ( - {\vec v _1}) = {\vec v _2} - {\vec v _1}$. If the time at ${{p}_1}$ is ${{t}_1}$ and at ${{p}_2}$ is ${{t}_2}$, the time interval during the motion along the curve between the two points is $\Delta t = {t_2} - {t_1}$. Now the average acceleration vector during the motion between the points is

${{\vec{a}}_{\text{av}}}=\frac{\Delta \vec v}{\Delta t} \tag{6} \label{6}$

When the time interval $\Delta t$ approaches zero the average acceleration in Eq. \eqref{6} becomes instantaneous acceleration. We call that instantaneous acceleration vector simply acceleration vector.

${{\vec{a}}_{\text{ins}}}=\vec{a}=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta \vec v}{\Delta t}=\frac{d\vec{v}}{dt}$

You know that the direction of the average acceleration vector is the same as the direction of $\Delta \vec v$. In Figure 8 notice that the direction of the average acceleration vector is directed (as indicated by the direction of $\Delta \vec v$) to the inside of the curved path.

Mechanics