Equations of motion (advanced derivations)

Dynamics is a branch of mechanics that deals with the motion of objects. It describes motion without considering the forces that cause the motion. One of the important components of dynamics is understanding the equations of motion. These equations describe the relationship between an object's velocity, acceleration, displacement, and time.

Understanding the basic terms

Before going into advanced derivations of the equations of motion, it is important to understand some basic terms used in kinematics:

• Displacement ((s)): It is the change in the position of an object. Displacement is a vector quantity, which means it has both magnitude and direction.
• Velocity ((v)): Rate of change of displacement. This is also a vector quantity and can be defined as (v = frac{Delta s}{Delta t}).
• Acceleration ((a)): Rate of change of velocity. It is a vector quantity expressed as (a = frac{Delta v}{Delta t}).
• Time ((t)): The measured period of time in which speed occurs or changes.

Derivation of equations of motion

The equations of motion are derived under the assumption that the acceleration ((a)) is constant. There are three fundamental equations of motion that must be understood:

First equation of motion

The first equation of motion is given as:

v = u + at

Where:

• (v) = terminal velocity
• (u) = initial velocity
• (a) = constant acceleration
• (t) = time

Derivative: Starting with the definition of acceleration, we have:

a = frac{v - u}{t}

Rearranging this equation gives:

v = u + at

Visual example

In the visual example, an object starts with an initial velocity (u) and over a period of time (t), it accelerates at a rate of (a) to reach a final velocity (v).

Second equation of motion

The second equation of motion is:

s = ut + frac{1}{2}at^2

Where:

• (s) = displacement

Derivation: Use the average velocity concept:

bar{v} = frac{u + v}{2}

Substitute (v) from the first equation of motion:

bar{v} = frac{u + (u + at)}{2} = u + frac{1}{2}at

Displacement ((s)) is obtained by multiplying the average velocity by time:

s = bar{v} times t = left(u + frac{1}{2}atright)t = ut + frac{1}{2}at^2

Visual example

In this visual example, an object starts from rest and accelerates at a rate of (a) in time (t), and travels a displacement (s).

Third equation of motion

The third equation of motion is:

v^2 = u^2 + 2as

Derivation: Start with the first two equations. From the first equation:

v = u + at

From the second, rewrite (t) as follows:

t = frac{v - u}{a}

Re-substitute this into the displacement expression:

s = ut + frac{1}{2}at^2 rightarrow s = uleft(frac{vu}{a}right) + frac{1}{2}aleft(frac{vu}{a}right)^2

Expand and rearrange, keeping (v^2) aside:

v^2 = u^2 + 2as

Visual example

In this view, an acceleration (a) is applied to the objects, resulting in the initial velocity (u) changing to a final velocity (v) over a displacement (s).

Application

Equations of motion are used in many applications in introductory level physics and engineering. They help calculate the trajectories of objects, understand forces in a system, and much more. Here are some examples:

Example 1: Freely falling objects

An object is dropped from a height of 100 m with an initial velocity of 0 m/s. Calculate the time it takes to fall to the ground.

Given: u = 0 m/s, s = 100 m, a = 9.8 m/s^2 (acceleration due to gravity).

Using s = ut + frac{1}{2}at^2 :

100 = 0 times t + frac{1}{2} times 9.8 times t^2

Solve for (t):

t^2 = frac{200}{9.8}

(t approx 4.52) sec

Example 2: Acceleration of a car

A car accelerates uniformly from rest to a speed of 25 m/s in 10 seconds. Find the distance covered by the car during this time.

Given: u = 0 m/s, v = 25 m/s, t = 10 s

Find (a) using the first equation of motion:

v = u + at rightarrow 25 = 0 + a times 10

(a = 2.5 m/s^2)

Now use the second equation of motion:

s = ut + frac{1}{2}at^2 = 0 times 10 + 0.5 times 2.5 times 100

(s = 125) meter

Example 3: Velocity of a thrown ball

A ball is thrown vertically upwards at a velocity of 20 m/s. Find the maximum height reached by the ball.

Given: u = 20 m/s , v = 0 m/s (at maximum height), a = -9.8 m/s^2

Using the third equation of motion:

v^2 = u^2 + 2as rightarrow 0 = 20^2 + 2 (-9.8) s

(400 = 19.6 sec)

Solve for (s):

(s approx 20.41) meters

Conclusion

Understanding the equations of motion provides an essential foundation for analyzing various motion types in physics. By mastering these equations, students can solve real-world problems and gain insight into the nature of motion. These equations are fundamental tools and remain important in advanced study and a variety of practical applications in physics and engineering.

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