What are displacement, velocity ,and acceleration?

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Displacement

Distance and displacement are two quantities that indicate the length between two points but they are completely different.

Distance means, the total length of the path covered by an object irrespective of direction and it is a scalar quantity. Displacement means the overall change in the position of an object or a particle. It is a vector quantity, which means it has both direction and magnitude.

Consider an object placed at A, it is displaced from its position to reach B. The displaced position of an object is known as displacement. Displacement is measured along a linear or straight line.  

Displacement, $\overrightarrow{AB} =\overrightarrow{OB}-\overrightarrow{ OA}$

It is the vectors differences between the beginning stage or initial position and the completion stage or final position. If A is the initial position and B is the final position. Then the displacement is,

$∆x=x_f-x_i$

where ∆x=displacement 

$x_f$= final position

$x_i$=initial position

The unit of displacement is meter (m).

Velocity

The rate of change of displacement of an object with respect to time and frame of reference is called velocity. It is a vector quantity.

$$\begin{gathered} Velocity=\frac{Displacement}{Time} \\ v=\frac{ds}{dt} \end{gathered}$$

where ds=change in displacement

dt= change in time

The unit of velocity is m/s or $m{s}^{-1}.$

Like distance and displacement, speed and velocity both have distinct meanings. Speed is a scalar quantity and it refers to how fast the object is moving or the rate at which the object covers the distance. If an object moves at a fast rate, it covers a large distance in small time. In contrast, if the object is moving slowly, that means slow speed, it covers only a small distance in a same time.

Instantaneous velocity

The definition for instantaneous velocity is the velocity of the object under motion at a specific point of time, which is almost zero.

$v_i=\underset{∆t\to \infty }{\lim }\frac{ds}{dt}$

Where $v_i$= instantaneous velocity

ds=change in displacement

dt= time interval

Average velocity

 The total displacement of the body divided by the total time taken is called average velocity. The formula is,

$v_{av}=\frac{d_2-d_1}{t_2-t_1}$

Where$v_{av}$= average velocity

$d_2 and d_1$= final and initial position.

$t_2 and t_1$=final and initial time.

Acceleration

It is described as the rate of change of velocity of an object with respect to time. It is a vector quantity. Also, if there is a change in the velocity of the object, then the object is said to accelerate. The formula is,

$$\begin{gathered} Acceleration =\frac{Change in velocity}{Time} \\ a=\frac{dv}{dt} \end{gathered}$$

The measuring unit of acceleration is $m/s^2 or m{s}^{-2}.$ 

 If the object's speed is rising at a constant rate, then the object has unvarying acceleration or constant acceleration.

If the ratio of velocity by time is evaluated for a finite interval of time, it is called average acceleration. If the ratio is evaluated for infinitesimally small interval time it is called instantaneous acceleration.

Acceleration in projectile motion

If an object is thrown upwards, the only primary force which acts on the object is gravity. Some other secondary forces are also acting on it, but compared to gravity, their effect is low. The force which drags the object downwards is called acceleration due to gravity. When an object is thrown obliquely from the earth's surface, it created a path that looks like a curve and reaches the center of the earth. The path created by the particle while traveling is called projectile and the motion is called projectile motion.

Equation of motion              

It is the so-called kinematic equations of motion. It also explains the relationship between displacement, velocity, and acceleration. It explains the motion of the particle which moves in 1- dimensional or 2- dimensional or 3- dimensional areas. The motion may be uniform or non-uniform, accelerating or non- accelerating. These equations relate the parameters and help to acknowledge the motion of the object. The four basic equations are,

$$\begin{gathered} v= u +at \\ s=ut+\frac{1}{2}a{t}^2 \\ {v}^2-u^2=2as \end{gathered}$$

Derivation

Let us imagine a particle that makes a displacement s in the time t. Let u is initial velocity and v is final velocity. The object's motion is uniformly accelerated in the time interval t at the rate of (a) in order to frame of reference. By using differential and integral calculus we can derive the equations.

First equation of motion

From the definition of acceleration, a is given by,

$a=\frac{dv}{dt}$

On integrating the equation, we get,

$$\begin{gathered} {\int }_u^{v}dv={\int }_0^{t}adt \\ {\left[v\right]}_u^v=a{\left[t\right]}_0^t \\ v-u=at \\ v=u+at \end{gathered}$$          (1)

Second equation of motion

The second equation of motion is derived from the velocity and from the first equation of motion.

$$\begin{gathered} v=\frac{ds}{dt} \\ ds= v dt \\ ds=(u+at)dt \end{gathered}$$

On integrating,

$$\begin{gathered} {\int }_0^{s}ds={\int }_0^t(u+at)dt \\ {\int }_0^{s}ds={\int }_0^{t}u dt+{\int }_0^{t}at dt \\ {\left[s\right]}_0^s=u{\left[t\right]}_0^t+a{\left[t\right]}_0^t \\ s=ut+\frac{1}{2}a{t}^2 \left(2\right) \end{gathered}$$    

Third equation of motion

The third equation is derived from the acceleration. It is given as,

$a=\frac{dv}{dt}$

Multiply and divide by ds on the right side,

$$\begin{gathered} a=\frac{dv}{dt}\frac{ds}{ds} \\ a=v\frac{dv}{ds} \\ vdv=a ds \end{gathered}$$

On integrating,

$$\begin{gathered} {\int }_u^{v}vdv={\int }_0^{s}ads \\ {\left[\frac{v^2}{2}\right]}_u^v=a{\left[s\right]}_0^s \\ \frac{v^2-u^2}{2}=as \\ {v}^2-u^2=2as \end{gathered}$$

Displacement, velocity, acceleration in simple harmonic motion

In classical mechanics, for 1-D SHM, the equation of motion is a second-order differential equation having constants coefficients. These constants are derived from Hooke's law and Newton's second law for a mass that is placed on a spring. The differential equation is,

$m \frac{d^{2}x}{d{t}^2}=-kx$      (5)

where m = inertial mass of the body.

x= displacement of the mass from the mean position.

k = the spring constant.

Hence,

$\frac{d^{2}x}{d{t}^2}=\frac{-k}{m}x$      (6)

on solving the above differential equation, we get a sinusoidal equation.

$x\left(t\right)=c_1\cos \left(\omega t\right)+c_2\sin \left(\omega t\right)$(7)

Where $\omega =\frac{k}{m}$

$c_1 and c_2$ are differential constants.

In order to find the values of these constants, put t=0,

$$\begin{gathered} x\left(0\right)=c_1\cos \left(0\right)+c_2\sin \left(0\right) \\ x\left(0\right)=c_1 \\ {c}_1=x_0 \end{gathered}$$

Where $c_1$ is the initial position of the particle.

By differentiating equation (7), we can find the constant c2,

$v\left(t\right)\equiv \frac{dx}{dt}=-\omega x_0\cos \omega t+c_2\omega \sin \left(\omega t\right)$

when t=0,

$$\begin{gathered} v\left(0\right)=\omega c_2 \\ {c}_2=\frac{v_0}{\omega } \end{gathered}$$

Substitute all the values in the equation (7),

$x\left(t\right)=x_0\cos \frac{k}{m}t+v_0\omega \sin \frac{k}{m}t$      (8)

The another form of the equation is,

$x\left(t\right)=A\cos (\omega t-\phi )$

Where$A=c_1^2+c_2^2$and$\tan \phi =\frac{c_2}{c_1}$ 

where A = amplitude.

ω=angular frequency.

φ= phase angle or initial phase. 

In this solution, the constants are determined from their initial conditions, and their original position is set in an equilibrium position. These two constants carry a physical value of this motion.

By using the calculus method, we can find the velocity and acceleration of the oscillation.

From the displacement equation, we can determine the velocity,

$$\begin{gathered} v\left(t\right)=\frac{dx}{dt}=\frac{d}{dt}\left(A\cos \right(\omega t-\phi \left) \\ v\right(t)=-A \sin (\omega t-\phi ) \end{gathered}$$

The first derivation of velocity is the acceleration. Therefore, by differentiating,

$$\begin{gathered} a\left(t\right)=\frac{dv}{dt}=\frac{d}{dt}(-A\sin (\omega t-\phi \left)\right) \\ a\left(t\right)=-A{\omega }^{2}c\mathrm{os}(\omega t-\phi ) \end{gathered}$$

The acceleration found from the above equation is at the equilibrium position. At the extreme points, the maximum acceleration is $A{\omega }^2$

It is interpreted that when a mass is subjected to oscillation, the acceleration is directly proportional to displacement.

a(x)=ωx

The graph shows the relationship between displacement, velocity, and acceleration in a simple harmonic motion. Whenever the velocity is maximum, the acceleration is zero.

The velocity is ahead by phase angle (π/2) from displacement. Acceleration is ahead by (π/2) from velocity or π from displacement.

Formulas

The formula of velocity is,

$v=\frac{dx}{dt}$

Average velocity formula is,

$v_{av}=\frac{d_2-d_1}{t_2-t_1}$

Instantaneous velocity formula is

$v_i=\underset{∆t\to \infty }{lim}\frac{ds}{dt}$

The formula of acceleration is,

$a=\frac{dv}{dt}$

The equations of motion are,

$$\begin{gathered} v= u +at \\ s=ut+\frac{1}{2}a{t}^2 \\ {v}^2-u^2=2as \end{gathered}$$

Context and Applications

It is a basic and necessary topic for all graduates and postgraduates particularly for, bachelors and masters in science (physics).

Practice Problems

Question 1: A car is traveling from rest. It attains a velocity of $20 m{s}^{-1}$ in 10 seconds. Find the acceleration of the car.

$$\begin{gathered} a. 2 m{s}^{-2} \\ b. 0.2 m{s}^{-2} \\ c. 4 m{s}^{-2} \\ d. 20 m{s}^{-2} \end{gathered}$$

Given data:

Initial velocity, u=$0 m{s}^{-1}$

Final velocity, v=$20 m{s}^{-1}$

Time, t=10s 

Explanation:

From the first equation of motion,

$$\begin{gathered} v-u=at \\ 20 m{s}^{-1}-0=a\times 10s \\ a=\frac{20 m{s}^{-1}}{10s} \\ a=2 m{s}^{-2} \end{gathered}$$

The acceleration of the car is $2 m{s}^{-2}$. 

Answer: The correct option is a.

Question 2: A ball is moving in a velocity 0.3m/s. Its velocity is decreasing at the rate of 0.05 $m/s^2$. Find the velocity of the ball after 5s.

$$\begin{gathered} a. 0.05 m{s}^{-1} \\ b. 0.5 m{s}^{-1} \\ c. 5 m{s}^{-1} \\ d. 50 m{s}^{-1} \end{gathered}$$

Given data:

Initial velocity, $u=0.3 m{s}^{-1}$

Acceleration, a=$0.05 m{s}^{-2}$

Time, t=5s

Solution:

By first kinematic equation,

$$\begin{gathered} v=u+at \\ v=(0.3 m{s}^{-1} )+(-0.05 m{s}^{-2} )\left(5s\right) \\ v=0.3 m{s}^{-1} -0.25 m{s}^{-1} \\ v=0.05 m{s}^{-1} \end{gathered}$$

The velocity of the ball is 0.05 m/s.

Answer: The correct option is b.

Question 3: A truck starts from rest with a uniform acceleration of $7 m/s^2$. Find the distance travelled by the truck after 5 s.

a. 75m b. 95m

c. 87.5m d. 75.5m

Given data:

Initial velocity, $u=0 m{s}^{-1}$

Acceleration, $a=7m{s}^{-2}$

Time, t=5 s

Solution:

From the second kinematic equation,

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ s=\left(0\right)\left(5s\right)+\frac{1}{2}\left(7m{s}^{-2}\right){\left(5s\right)}^2 \\ s=87.5 m \end{gathered}$$

The distance travelled by the truck is 87.5 m.

Answer: The correct option is c.

Question 4: Average velocity is______.

a. Total distance/time taken

b. Total displacement/total time taken

c. Distance/speed

d. Speed/time

Answer: The correct option is (b).

Explanation: The average velocity is defined as the total displacement of the body divided by total time taken.

             $v_{av}=\frac{∆d}{∆t}=\frac{d_2-d_1}{t_2-t_1}$

Question 5: The motion of the object from one position to another position is called ________.

a. Velocity

b. Speed

c. Displacement

d. Distance

Answer: The correct option is (c).

Explanation: Displacement is defined as the movement of an object from one place to another. And it is the shortest distance between the object's initial position and final position.