Derivation Of Equation Of Motion

Deriving the Equations of Motion: A Comprehensive Guide

Understanding how objects move is fundamental to physics. This article delves into the derivation of the equations of motion, crucial tools for analyzing the motion of objects under constant acceleration. We'll explore the underlying principles, the mathematical derivations, and their practical applications. This guide is designed to be accessible to students of all backgrounds, building a strong foundation for understanding more complex physics concepts. By the end, you'll not only understand how to use these equations, but also why they work.

Introduction: Understanding Motion and Acceleration

Before diving into the derivations, let's establish a common understanding of basic kinematic quantities. We're primarily concerned with objects moving in one dimension (along a straight line). Our key players are:

• Displacement (s): The change in an object's position. It's a vector quantity, meaning it has both magnitude (how far) and direction. In one dimension, we often represent it simply as the distance from a starting point. Units: meters (m).
• Velocity (v): The rate of change of displacement. It's also a vector quantity. Average velocity is the total displacement divided by the total time taken. Instantaneous velocity is the velocity at a specific instant in time. Units: meters per second (m/s).
• Acceleration (a): The rate of change of velocity. Like displacement and velocity, it’s a vector quantity. Average acceleration is the change in velocity divided by the time taken. Instantaneous acceleration is the acceleration at a specific instant in time. We'll focus primarily on constant acceleration in this derivation. Units: meters per second squared (m/s²).
• Time (t): The duration of the motion. Units: seconds (s).

The equations of motion allow us to relate these four quantities, enabling us to predict the future position or velocity of an object given its initial conditions and acceleration.

Derivation of the First Equation of Motion: v = u + at

This equation relates final velocity (v) to initial velocity (u), acceleration (a), and time (t). Let's derive it using the definition of acceleration:

Acceleration is the rate of change of velocity:

a = (v - u) / t

where:

• 'v' is the final velocity
• 'u' is the initial velocity
• 't' is the time taken

To obtain the first equation of motion, we simply rearrange this equation to solve for 'v':

v = u + at

This equation assumes constant acceleration. If acceleration is not constant, this equation would only provide an approximation over small time intervals.

Derivation of the Second Equation of Motion: s = ut + ½at²

This equation connects displacement (s) with initial velocity (u), acceleration (a), and time (t). We'll derive it using the concept of average velocity.

Average velocity is the average of initial and final velocities when acceleration is constant:

Average velocity = (u + v) / 2

We know that average velocity is also equal to total displacement divided by total time:

Average velocity = s / t

Equating these two expressions for average velocity:

s / t = (u + v) / 2

Now, substitute the first equation of motion (v = u + at) into this equation:

s / t = (u + u + at) / 2

s / t = (2u + at) / 2

Multiplying both sides by 't':

s = ut + ½at²

This is our second equation of motion. It provides a direct relationship between displacement and time, given constant acceleration.

Derivation of the Third Equation of Motion: v² = u² + 2as

This equation doesn't explicitly involve time. We derive it by combining the first and second equations of motion.

From the first equation of motion:

t = (v - u) / a

Substitute this expression for 't' into the second equation of motion:

s = u[(v - u) / a] + ½a[(v - u) / a]²

Simplifying this equation:

s = (uv - u²) / a + (v² - 2uv + u²) / (2a)

Multiplying both sides by 2a:

2as = 2uv - 2u² + v² - 2uv + u²

Further simplifying:

2as = v² - u²

Rearranging to obtain the third equation of motion:

v² = u² + 2as

This equation is particularly useful when time is unknown or not relevant to the problem.

Derivation for Objects Under Gravity (Free Fall)

When dealing with objects falling freely under the influence of gravity, we can adapt the equations. The acceleration due to gravity (g) is approximately 9.81 m/s² downwards. We usually take the downwards direction as positive. Therefore, the equations become:

• v = u + gt
• s = ut + ½gt²
• v² = u² + 2gs

These equations are extremely useful for analyzing projectile motion, a topic that builds upon these fundamental concepts.

Limitations and Assumptions

It’s crucial to remember that these equations rely on several key assumptions:

• Constant acceleration: The acceleration of the object remains constant throughout the motion. In real-world scenarios, this is often an approximation. Air resistance, for example, can significantly affect acceleration.
• One-dimensional motion: The motion is along a straight line. For two or three-dimensional motion, vector calculus is required.
• Negligible air resistance: Air resistance is ignored in these derivations. For high-speed objects or objects with large surface areas, air resistance can significantly impact the motion.

Solving Problems Using the Equations of Motion

Let's illustrate the application of these equations with an example:

Problem: A car accelerates uniformly from rest to 20 m/s in 5 seconds. Calculate:

a) its acceleration b) the distance it travels during this time.

Solution:

a) We can use the first equation of motion, v = u + at:

• v = 20 m/s
• u = 0 m/s (starts from rest)
• t = 5 s

Rearranging to solve for 'a':

a = (v - u) / t = (20 - 0) / 5 = 4 m/s²

b) Now, we can use the second equation of motion, s = ut + ½at²:

• u = 0 m/s
• a = 4 m/s²
• t = 5 s

Substituting these values:

s = 0(5) + ½(4)(5)² = 50 m

Therefore, the car's acceleration is 4 m/s², and it travels 50 meters during the 5 seconds.

Frequently Asked Questions (FAQ)

Q1: What if acceleration is not constant?

A1: For non-constant acceleration, the equations of motion we’ve derived are not directly applicable. Calculus (integration) is necessary to analyze such motions. You would need to know the acceleration as a function of time (a(t)) and integrate to find velocity and displacement.

Q2: How do these equations apply to vertical motion?

A2: For vertical motion under gravity, substitute 'g' (acceleration due to gravity) for 'a' in the equations, remembering to consider the direction of gravity (usually downwards as positive).

Q3: Can these equations be used for projectile motion?

A3: These equations can be applied to the vertical and horizontal components of projectile motion separately, as long as air resistance is neglected. The horizontal component usually has a constant velocity (zero acceleration), while the vertical component has constant acceleration due to gravity.

Q4: What is the difference between speed and velocity?

A4: Speed is a scalar quantity (magnitude only), while velocity is a vector quantity (magnitude and direction).

Conclusion

The derivation and application of the equations of motion are fundamental to understanding classical mechanics. Mastering these equations provides a strong foundation for exploring more advanced topics in physics, such as projectile motion, collisions, and rotational dynamics. Remember the underlying assumptions – constant acceleration and one-dimensional motion – and you'll be well-equipped to tackle a wide range of motion problems. By understanding the derivations, you'll not only be able to solve problems efficiently but also gain a deeper appreciation for the elegance and power of fundamental physics principles. Remember to always carefully analyze the problem statement, identify the known quantities, and select the appropriate equation before plugging in the values and solving for the unknown. Practice is key to mastering these crucial tools for understanding the world around us.