The Derivation and Application of Kinematic Equations
- You're standing at a traffic light, watching a car accelerate smoothly from rest.
- Its velocity increases steadily, and you start to wonder: how far has it traveled after 5 seconds? How fast is it moving now?
These are the kinds of questions that the kinematic equations can help you answer.
Deriving the Kinematic Equations for Uniformly Accelerated Motion
We consider an object moving on a straight line, with constant acceleration. Let’s derive its velocity and displacement step by step.
1. Relating Velocity and Time
- Consider an object starting with an initial speed ($u$) and accelerating at a constant rate ($a$) for some time t until reaching a velocity $v(t)$ at time $t$.
- How does its velocity change over time?
- What is the form of $v(t)$ ?
- Acceleration is defined as the rate of change of velocity:
$$a = \frac{\Delta v}{\Delta t} \quad \implies a = \frac{v(t)-u}{t} \implies \quad v(t) = u + at$$
This is the equation, which relates states how velocity $v$ changes as a function of time $t$.Analogy
- Think of acceleration as a steady push on a swing.
- The longer and stronger you push (time), the faster the swing moves (velocity).
A ball is thrown vertically upward with an initial velocity of $u = 15 \, \text{m/s}$. Assuming $a = -9.8 \, \text{m/s}^2$ (acceleration due to gravity), find its velocity after $t = 2 \, \text{s}$.
Solution
Using $v = u + at$:
$$v = 15 + (-9.8)(2)$$
$$v = 15 - 19.6 = -4.6 \, \text{m/s}$$
The negative sign indicates that the ball is moving downward after 2 seconds.
2. Relating Displacement and Time
Now, let’s figure out how far the object travels (displacement, $s$) in a given time. To do that, let's approach the problem graphically.
- From previous section, we know, that displacement is actually the area under velocity-time graph.
- We have also established the velocity-time dependence as a linear function $v(t)$.
- In the following, we consider motion with positive projection of initial velocity and acceleration on the axis of motion, but the approach is general enough to fit any of the other cases.

We see that essentially the problem of finding the area under this linear function reduces to finding the area of trapezium or, even simpler, of rectangle and right-angle triangle. We obtain:
$$\text{Area} = s(t) = u t + \frac{1}{2}(v(t)-u) t $$
$$ = ut + \frac{1}{2}(u+at-u) t $$
$$= ut + \frac{1}{2}at^2$$
This is the equation, which relates displacement to time in uniformly accelerated motion.
A car starts from rest ($u = 0$) and accelerates uniformly at $a = 2 \, \text{ m s}^{-2}$ for $t = 5 \text{ s}$. How far does it travel?
Solution
Using $s = ut + \frac{1}{2}at^2$:
$$s = 0 \cdot 5 + \frac{1}{2}(2)(5^2)$$
$$s = \frac{1}{2}(2)(25) = 25 \, \text{m}$$
The car travels 25 meters.
3. Relating Velocity and Displacement
Finally, let’s connect velocity and displacement without involving time.
- We start with two standard equations: $$v = u + at$$ $$s = ut + \frac{1}{2}at^2$$
- Our goal is to eliminate time $t$.
- We first express $t$ from first equation: $$t = \frac{v - u}{a}$$
- Replace $t$ in the displacement formula: $$s = u \left( \frac{v - u}{a} \right) + \frac{1}{2}a \left( \frac{v - u}{a} \right)^2$$
- Now, we simplify each term: $$s = \frac{u(v - u)}{a} + \frac{(v - u)^2}{2a}$$


