Kinematics Equations Explained: Formulas, Derivations and Examples

What Is Kinematics

When we observe nature, we see many objects in motion. From experience, we recognize that motion represents the continuous change in the position of an object with time. We study dynamics in two parts: kinematics and kinetics.

In kinematics, we study motion in terms of space and time without knowing the cause of motion (the forces that caused the motion). In kinetics, we are also concerned about the cause of motion (forces). Students working through these concepts in broader engineering contexts may find it useful to connect with an electrical engineering tutor for related coursework in circuits and electromagnetism.

Kinematic Equations List

${v_x} = {v_{ox}} + {a_x}t$ …(1)

$x = {x_0} + {v_{ox}}t + frac{1}{2}{a_x}{t^2}$ …(2)

$v_x^2 = v_{ox}^2 + 2{a_x}(x – {x_0})$ …(3)

$(x – {x_0}) = frac{1}{2}left( {{v_{ox}} + {v_x}} right)t$ …(4)

Equations (1), (2) & (3) (and sometimes (4)) are the equations of motion with constant acceleration. These are also known as equations of kinematics or kinematic equations.

${v_x}:$ velocity of particle along x-axis at time t

${v_{ox}}:$ velocity of particle along x-axis at t = 0

${a_x}:$ constant acceleration of particle along x-axis

x: position of the particle at time t on the x-axis

${x_o}:$ position of particle at time t = 0 on the x-axis

t: it represents general time.

How to Use Kinematic Equations

We have the first equation of kinematics ${v_x} = {v_{ox}} + {a_x}t$. This expression enables us to determine the velocity at any time t if the initial velocity, the acceleration, and the elapsed time are given.

We have the second equation of kinematics $x = {x_o} + {v_{ox}}t + frac{1}{2}{a_x}{t^2}$. This expression enables us to determine the distance traveled during any time t if the initial velocity, acceleration, and time elapsed are known.

We have a third equation that lets us determine the final velocity if the initial velocity, acceleration, and displacement are known. Those studying related applied mechanics will find these principles also underpin structural analysis coursework.

Solved Problems on Each Kinematic Equation

Example 1

An athlete takes 20 sec to reach his maximum speed of 18.0 km/h. What is the magnitude of average acceleration?

Ans. The athlete is running along the x-axis with constant acceleration; we can use

${v_x} = {v_{ox}} + {a_x}t$

$18 times frac{5}{18},text{m/s} = 0 + {a_x} times 2,text{sec}$

$5,text{m/s} = {a_x} times 2,text{sec}$

${a_x} = 2.5,text{m/s}^2$

Therefore, the magnitude of the average acceleration is 2.5 m/s².

Example 2

An object having a velocity of 4.0 m/s is accelerated at the rate of 1.2 m/s² for 5.0 sec. Find the distance traveled during the period of acceleration.

Ans. Suppose the object is moving along the +ve x-axis. Assuming that the initial location of the particle is at the origin.

Here, $x – {x_o} = x – 0 = x$

${v_{ox}} = 4.0,text{m/s}$, ${a_x} = 1.2,text{m/s}^2$, $t = 5,text{sec}$.

Using kinematics equation,

$x = {x_o} + {v_{ox}}t + frac{1}{2}{a_x}{t^2}$

$Rightarrow x – {x_o} = {v_{ox}}t + frac{1}{2}{a_x}{t^2}$

$Rightarrow x = left[ {4 times 5 + frac{1}{2} times 1.2 times {(5)}^2} right]text{meter}$

$Rightarrow x = left[ {20 + frac{1.2}{2} times 25} right]text{meter}$

$Rightarrow x = 35,text{meter}$

Example 3

A person traveling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s² to his scooter. How far will it travel before stopping?

Ans. Suppose the person is traveling along the positive x-axis and initially that person was at the origin (x = 0). He is moving along a straight line with constant negative acceleration.

Using kinematics equation,

$v_x^2 = v_{ox}^2 + 2{a_x}(x – {x_o})$

Here, ${v_x} = 0$, ${v_{ox}} = 12,text{m/s}$, ${a_x} = -6.0,text{m/s}^2$, $(x – {x_o}) = ?$

$Rightarrow quad {0^2} = {(12,text{m/s})^2} + 2(-6,text{m/s}^2)(x)$ [because $x – {x_o} = x – 0 = x$]

$Rightarrow quad 12x = 144$

$Rightarrow quad x = 12,text{m}$

So, the person will travel a distance of 12 m before stopping.

Kinematic Equations for Projectile Motion

The projectile’s motion can be discussed separately for the horizontal and vertical parts. We take origin $({x_o} = 0,,{y_o} = 0)$ at the point of projection. The instant when the particle was projected is taken as t = 0.

The plane of motion is taken as the x-y plane, the horizontal line OX as the x-axis and the vertical line OY as the y-axis. Vertically upward direction is taken as the positive direction of the y-axis.

We have, ${v_{ox}} = {v_o}cos{alpha_o}$, ${a_x} = 0$

${v_{oy}} = {v_o}sin{alpha_o}$, ${a_y} = -g$

Horizontal motion

${a_x} = 0$, we have

${v_x} = {v_{ox}} + {a_x}t$

$Rightarrow quad {v_x} = {v_{ox}}$

Again, $x – {x_0} = {v_{ox}}t + frac{1}{2}{a_x}{t^2}$ [Here, ${a_x} = 0$, ${v_{ox}} = {v_o}sin{alpha_o}$, ${x_o} = 0$]

$Rightarrow quad x – {x_0} = {v_{ox}}t$

$Rightarrow quad x = {x_0} + {v_{ox}}t$

Vertical Motion

The acceleration of the particle is g in the downward direction. Thus ${a_y} = -g$ [Here, ${a_y} = -g$, ${v_{oy}} = {v_o}sin{alpha_o}$, ${y_o} = 0$]

We can write, ${v_y} = {v_{oy}} + {a_y}t$

$Rightarrow {v_y} = {v_{oy}} – gt$

and $y – {y_o} = {v_{oy}}t + frac{1}{2}{a_y}{t^2}$

$Rightarrow y = {y_o} + {v_{oy}}t + frac{1}{2}{a_y}{t^2}$

Again, $v_y^2 = v_{oy}^2 + 2{a_y}(y – {y_o})$

Kinematic Equations for Free Fall

A typical example of motion in a straight line with constant acceleration is the free fall of a body near the earth’s surface.

Here, we have taken the origin O at the starting point and the upward direction as positive. Assuming the initial position coordinate ${y_o}$ and initial velocity is ${v_{oy}}$.

The y-acceleration is downward (i.e., along the –ve y-direction) therefore, ${a_y} = -g$.

After time t, its position coordinate can be written as

$y – {y_o} = {v_{oy}}t + frac{1}{2}{a_y}{t^2}$

$Rightarrow y = {y_o} + {v_{oy}}t + frac{1}{2}(-g){t^2}$

Here in this case, ${y_o} = 0$, ${v_{oy}} = 0$

After time t, its velocity coordinate ${v_y} = {v_{oy}} + {a_y}t$

$Rightarrow {v_y} = {v_{oy}} – gt$

Here ${v_{oy}} = 0$ and ${a_y} = -g$

We can also write, $v_y^2 = v_{oy}^2 + 2{a_y}(y – {y_o})$

$Rightarrow v_y^2 = v_{oy}^2 + 2(-g)(y – {y_o})$

Here, ${a_y} = -g$ and ${y_o} = 0$. Students in life-sciences programs who encounter free-fall and motion concepts in biomechanics may benefit from working with a biomedical engineering tutor for deeper applied context.

Kinematic Equations for Rotation

When a body revolves about a stationary axis (usually z-axis), the position of the body is described by an angular position coordinate $theta$. If the angular acceleration is constant, then $theta$, ${omega_2}$ and ${alpha_2}$ are related by simple kinematic equations.

Kinematic equations in the case of rotation are as follows:

$theta – {theta_o} = {omega_{oz}} + frac{1}{2}{alpha_z}{t^2}$

$Rightarrow {omega_z} = {omega_{oz}} + {alpha_z}t$

$Rightarrow omega_z^2 = omega_{oz}^2 + 2{alpha_z}(theta – {theta_o})$

Here, ${theta_o}:$ Angular position of the body at time t = 0

$theta:$ Angular position of the body at time t = t

${alpha_z}:$ Constant angular acceleration of the body

${omega_{oz}}:$ Angular velocity of the body at time t = 0

${omega_z}:$ Angular velocity of the body at time t = t

Those preparing for standardized engineering exams that include rotational dynamics can find structured support through test preparation tutoring.

Kinematic Equation Derivation Using Calculus

Let a particle be moving along the x-axis. Initially, the particle was at $x = {x_o}$ with initial velocity ${v_{ox}}$.

The particle has a constant acceleration along the x-axis. From the definition of acceleration,

${a_x} = frac{d{v_x}}{dt} quad Rightarrow quad d{v_x} = {a_x},dt$

Integrating both sides with proper limits,

$intlimits_{{v_{ox}}}^{{v_x}} {d{v_x}} = intlimits_0^t {{a_x}},dt$

$Rightarrow {v_x} – {v_{ox}} = {a_x}(t – 0)$

$Rightarrow quad {v_x} = {v_{ox}} + {a_x}t$

Again from the definition of velocity,

${v_x} = frac{dx}{dt} Rightarrow frac{dx}{dt} = {v_{ox}} + {a_x}t$

$Rightarrow dx = ({v_{ox}} + {a_x}t),dt$

Now, integrating both sides with proper limits,

$intlimits_{{x_o}}^x {dx} = intlimits_0^t {({v_{ox}} + {a_x}t),dt}$

$Rightarrow x – {x_o} = {v_{ox}}(t – 0) + {a_x}left[ frac{t^2}{2} right]_0^t$

$Rightarrow quad x – {x_o} = {v_{ox}}t + frac{1}{2}{a_x}{t^2}$

Again starting from, ${a_x} = frac{d{v_x}}{dt}$

$frac{d{v_x}}{dx} cdot frac{dx}{dt} = {a_x} Rightarrow {v_x} cdot frac{d{v_x}}{dt} = {a_x}$

$Rightarrow {v_x} cdot d{v_x} = {a_x}(dx)$

Now, integrating both sides,

$intlimits_{{v_{ox}}}^{{v_x}} {{v_x},dx} = intlimits_{{x_o}}^x {{a_x}(dx)} Rightarrow left[ frac{v_x^2}{2} right]_{{v_{ox}}}^{{v_x}} = {a_x}(x – {x_o})$

$Rightarrow quad v_x^2 = v_{ox}^2 + 2a_x(x – {x_o})$

Students who find calculus-based derivations challenging alongside their chemical engineering coursework can get support from a chemical engineering tutor for the broader mathematical methods involved.

Example

A particle starts with an initial velocity of 2.5 m/s along the positive x-direction and accelerates uniformly at the rate of 0.5 m/s². Find the distance traveled by it in the first two seconds.

Ans. Let the particle start its motion from the origin with an initial velocity of ${v_{ox}} = 2.5,text{m/s}$, ${a_x} = 0.5,text{m/s}^2$

Using, ${a_x} = frac{d{v_x}}{dt} Rightarrow d{v_x} = {a_x},dt$

Now, integrating both sides,

$intlimits_{{v_{ox}}}^{{v_x}} {d{v_x}} = {a_x}intlimits_0^t {dt}$

$Rightarrow {v_x} – 2.5 = frac{1}{2}(t – 0) Rightarrow {v_x} = 2.5 + frac{t}{2}$

Again using ${v_x} = frac{dx}{dt}$

$Rightarrow frac{dx}{dt} = 2.5 + frac{t}{2} Rightarrow dx = left( 2.5 + frac{t}{2} right)dt$

Now, integrating both sides,

$intlimits_{{x_o}}^x {dx} = intlimits_0^2 {left( 2.5 + frac{t}{2} right)dt}$

$Rightarrow x – {x_o} = 2.5(2 – 0) + frac{1}{2}frac{(2)^2}{2}$

Since the particle does not turn back, the distance traveled equals 6 m.

Kinematics Equations for Nonuniform Acceleration

If the acceleration of a particle during its motion does not remain constant, then both motion and acceleration are nonuniform.

As by definition, ${v_x} = frac{dx}{dt}$, ${a_x} = frac{d^2x}{dt^2}$

If x is a function of time, the second derivative of displacement with respect to time represents acceleration. If velocity is a function of position, then by chain rule of differentiation,

${a_x} = frac{d{v_x}}{dt} = frac{d{v_x}}{dx} cdot frac{dx}{dt}$

$Rightarrow {a_x} = vfrac{d{v_x}}{dx}$

For students exploring how these differential methods connect to real-world language and communication modeling, our overview of communication and media studies tutoring shows how analytical thinking transfers across disciplines.

Example

The acceleration ${a_x}$ of a particle moving in a straight line varies with its displacement $x = 2{a_x}$. The velocity of the particle is zero at zero displacement. Find the corresponding velocity displacement equation.

Ans. We have, ${a_x} = 2x$

$Rightarrow frac{d{v_x}}{dx} cdot frac{dx}{dt} = 2x$

$Rightarrow {v_x}(d{v_x}) = 2x(dx)$

Now, integrating both sides with proper limits,

$intlimits_0^{{v_x}} {{v_x}(d{v_x})} = intlimits_0^x {2x(dx)}$

$Rightarrow frac{v_x^2}{2} = 2 cdot frac{x^2}{2} Rightarrow v_x^2 = 2x^2$

$Rightarrow quad {v_x} = pmsqrt{2},x$

Resources

Kinematics Equations Calculator

While there are many online calculators on kinematics equations, we suggest you use Physics Catalyst’s calculator on the same.

Online Tutoring in Kinematics Equations and Related Physics Topics

My Engineering Buddy provides online physics tutoring that covers topics like kinematics equations and much more.

Related: Motion in two dimensions 2d in physics | Significant figures | Physics Homework Help | Physics Tutor Online

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