Derivation of Formulas for Uniformly Accelerated Linear Motion

In the study of classical mechanics, Uniformly Accelerated Rectilinear Motion (UARM) represents one of the most fundamen

In the study of classical mechanics, Uniformly Accelerated Rectilinear Motion (UARM) represents one of the most fundamental types of movement. Whether analyzing the descent of an object under gravity or the acceleration of a vehicle on a straight track, understanding the mathematical laws governing this motion is essential for solving more complex dynamical problems in physics and engineering.
To begin our derivation, we must establish a precise definition. An object is said to undergo uniformly accelerated rectilinear motion if it moves along a straight line and its acceleration remains constant over time. This implies that the velocity of the object changes by equal amounts during equal intervals of time.

To maintain mathematical rigor, we define the following physical quantities and symbols:

  • $t$: The elapsed time (s)
  • $x_0$: The initial position (m)
  • $x$: The final position at time $t$ (m)
  • $\Delta x$: The displacement, defined as $x - x_0$ (m)
  • $v_0$: The initial velocity (m/s)
  • $v$: The final velocity at time $t$ (m/s)
  • $a$: The constant acceleration ($\text{m/s}^2$)

1. The Velocity-Time Relationship

The concept of acceleration is rooted in the rate of change of velocity. By definition, acceleration is the change in velocity per unit of time:

$$a = \frac{\Delta v}{\Delta t} = \frac{v - v_0}{t}$$

To find the expression for the final velocity, we can use simple algebraic rearrangement. By multiplying both sides by $t$ and adding $v_0$ to both sides, we arrive at the first kinematic equation:

$$v = v_0 + at$$

Physical Insight: This formula demonstrates a linear relationship between velocity and time. If $a > 0$, the velocity increases over time (acceleration); if $a < 0$, the velocity decreases (deceleration or retardation).

2. The Displacement-Time Relationship

Determining how far an object has traveled requires a different approach. In UARM, because the velocity changes at a constant rate, the average velocity ($\bar{v}$) over a time interval $t$ is simply the arithmetic mean of the initial and final velocities:

$$\bar{v} = \frac{v_0 + v}{2}$$

Since displacement is the product of average velocity and time ($\Delta x = \bar{v} \cdot t$), we can substitute the average velocity expression into this definition:

$$\Delta x = \left( \frac{v_0 + v}{2} \right) t$$

To express this solely in terms of acceleration and time, we substitute the velocity formula derived in the previous section ($v = v_0 + at$) into the equation above:

$$\Delta x = \frac{v_0 + (v_0 + at)}{2} \cdot t$$

$$\Delta x = \frac{2v_0 + at}{2} \cdot t$$

Distributing the $t$ and simplifying the fraction, we obtain the second kinematic equation:

$$\Delta x = v_0 t + \frac{1}{2} at^2$$

If we assume the starting position $x_0 = 0$, the position $x$ at any time $t$ is given by $x = v_0 t + \frac{1}{2} at^2$.

3. The Velocity-Displacement Relationship

In many practical scenarios, the time elapsed is unknown, but the displacement and velocities are given. To solve these problems, we need a "timeless" equation. We can derive this by eliminating the time variable $t$ from our previous equations.

First, we rearrange the velocity formula to solve for $t$:

$$t = \frac{v - v_0}{a}$$

Next, we substitute this expression for $t$ into the displacement formula $\Delta x = \frac{v_0 + v}{2} t$:

$$\Delta x = \left( \frac{v + v_0}{2} \right) \left( \frac{v - v_0}{a} \right)$$

Applying the algebraic identity for the difference of squares, $(v + v_0)(v - v_0) = v^2 - v_0^2$, the equation becomes:

$$\Delta x = \frac{v^2 - v_0^2}{2a}$$

Rearranging this to isolate the velocity terms gives us the third kinematic equation:

$$v^2 - v_0^2 = 2a \Delta x$$

This equation is an extremely powerful tool in kinematics, allowing for the direct calculation of velocity or displacement without requiring a temporal measurement.

Practical Application Example

To illustrate the utility of these formulas, let us consider a real-world scenario.

Problem:
A car is traveling along a straight highway with an initial velocity of $10\text{ m/s}$. The driver presses the accelerator, resulting in a constant acceleration of $2\text{ m/s}^2$. Calculate:

  1. The velocity of the car after $3\text{ seconds}$.
  2. The total displacement covered during these $3\text{ seconds}$.

Solution:

Given:

  • $v_0 = 10\text{ m/s}$
  • $a = 2\text{ m/s}^2$
  • $t = 3\text{ s}$

1. Finding the final velocity ($v$):
Using the first kinematic equation:
$$v = v_0 + at$$
$$v = 10 + (2 \times 3) = 16\text{ m/s}$$

2. Finding the displacement ($\Delta x$):
Using the second kinematic equation:
$$\Delta x = v_0 t + \frac{1}{2} at^2$$
$$\Delta x = (10 \times 3) + \frac{1}{2}(2)(3^2)$$
$$\Delta x = 30 + 9 = 39\text{ m}$$

Conclusion

The derivation of these three formulas provides a complete mathematical framework for describing motion under constant acceleration. By mastering these relationships, one can transition from simple observations of movement to rigorous quantitative analysis, laying the groundwork for more advanced studies in dynamics and Newton's Laws of Motion.