Grade 10

Grade 10MechanicsDynamics


Equations of motion


In physics, the equations of motion are used to describe the behavior of a moving object. These equations provide a way to calculate key characteristics of motion such as displacement, velocity, and acceleration. They are important for understanding how objects move and are fundamental concepts in the study of mechanics. In this lesson, we will explore these equations, their derivations, and how they can be applied to solve problems in dynamics.

Introduction to kinematics and motion

Dynamics is the study of motion without considering the forces that cause the motion. It focuses on the position, velocity and acceleration of objects. In dynamics, we describe motion in terms of:

  • Displacement - The change in the position of an object. It is a vector quantity, meaning it has both magnitude and direction.
  • Velocity - The rate of change of displacement. Like displacement, velocity is a vector and is usually expressed in meters per second (m/s).
  • Acceleration - The rate of change in velocity. It shows how the velocity of an object changes over time and is measured in meters per second squared (m/s²).

Motion can be uniform or non-uniform. In uniform motion, the velocity remains constant, which means zero acceleration. In non-uniform motion, the velocity changes and hence the acceleration is not zero.

Three equations of motion

The three equations of motion are derived based on the assumptions of uniformly accelerated motion, which means that the acceleration remains constant over a time interval. The three equations are fundamental to solving many physics problems related to motion and can be described as:

First equation of motion

The first equation of motion relates velocity with respect to time. It can be stated as:

v = u + at

Where:

  • v is the final velocity of the object.
  • u is the initial velocity of the object.
  • a is the constant acceleration.
  • t is the time taken.

Example 1

Consider an object starting from rest, which means initial velocity u = 0. It accelerates at a constant rate of 5 m/s² for 3 seconds. Using the first equation of motion:

v = u + at
v = 0 + (5 m/s² * 3 s)
v = 15 m/s

The final velocity of the object is 15 m/s.

Visual explanation (Equation 1)

u = 0 a = 5 m/s² v = 15 m/s

Second equation of motion

The second equation of motion relates displacement to time, initial velocity and acceleration. It is given as:

s = ut + (1/2)at²

Where:

  • s is the displacement.
  • u, a, and t are the same as those defined previously.

Example 2

Using the previous example, where u = 0 m/s, a = 5 m/s², and t = 3 s, calculate the displacement:

s = ut + (1/2)at²
s = 0*3 + 0.5*(5 * 3²)
s = 0.5 * 5 * 9
s = 22.5 m

The displacement is 22.5 metres.

Visual explanation (Equation 2)

S = 22.5 m

Third equation of motion

The third equation of motion provides a relation that is independent of time. It connects the final velocity with the initial velocity, acceleration, and displacement:

v² = u² + 2as

Where:

  • v and u are the final and initial velocities, respectively.
  • a is the constant acceleration.
  • s is the displacement.

Example 3

Suppose an object accelerates from rest to a velocity of 15 m/s² with a constant acceleration of 5 m/s². Find the displacement.

v² = u² + 2as
15² = 0² + 2*5*s
225 = 10s
s = 225 / 10
s = 22.5 m

The calculated displacement is again 22.5 m, which is consistent with our previous calculation.

Visual explanation (Equation 3)

u = 0 m/s v = 15 m/s S = 22.5 m

Application and understanding

These equations allow us to make predictions about motion based on initial conditions and known quantities such as time or acceleration. They are fundamental in fields such as engineering, aerodynamics, the automotive industry, sports science, and wherever an understanding of objects in motion is essential.

Let's consider some additional examples that cover different scenarios for applying these equations.

Example 4: Free-fall

If an object is dropped from a height and falls under the effect of gravity only, it experiences a constant acceleration due to gravity, approximately g = 9.8 m/s². Calculate the velocity and displacement of a ball dropped from a 45 m high building after 3 seconds.

Using the first equation of motion:

v = u + at
u = 0, a = 9.8 m/s², t = 3 s
v = 0 + (9.8 * 3)
v = 29.4 m/s

The velocity after 3 seconds is 29.4 m/s.

Using the second equation of motion for displacement:

s = ut + (1/2)at²
s = 0*3 + 0.5*(9.8 * 3²)
s = 0.5 * 9.8 * 9
s = 44.1 m

The ball falls 44.1 m in 3 sec.

Example 5: Projectile motion

Consider a projectile launched with velocity u at an angle θ to the horizontal. Horizontally, the acceleration is zero, while vertically, the acceleration is g. We can divide this motion into two parts: horizontal and vertical.

The velocity of the horizontal motion remains constant because there is no horizontal acceleration:

x = u*cos(θ)*t

Vertical motion is affected by gravity:

y = u*sin(θ)*t - 0.5*g*t²

Here x and y represent the position at any time, which helps in tracking the projectile path.

Understanding these equations through various scenarios highlights their utility and importance as a fundamental part of physics education. By understanding these concepts, one can analyze and predict the motion of objects in the physical world.

Conclusion

The equations of motion enable students to solve real-world problems involving anything from simple free-fall motion to complex trajectories. Mastering these equations provides the tools necessary to delve further into advanced topics in physics such as dynamics, fluid mechanics, and astrophysics.

Practical application of these equations using examples as well as visual aids enhances the understanding and retention of these fundamental concepts, which form an important part of the basic physics course.


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