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 $t$.
- 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: $$s = \frac{u(v - u)}{a} + \frac{(v - u)^2}{2a}$$
- And combine over the common denominator: $$s = \frac{2u(v - u) + (v - u)^2}{2a}$$
- Factor the numerator:$$s = \frac{(v - u)[2u + (v - u)]}{2a} = \frac{(v - u)(v + u)}{2a}$$
- Now use the identity \((v - u)(v + u) = v^2 - u^2\):
$$s = \frac{v^2 - u^2}{2a}$$ - Finally, rearrange: $$v^2 = u^2 + 2as$$
This equation directly links the initial and final speeds of an object with its displacement under constant acceleration - without needing to know the time.
Summary of Kinematic Equations
- Here’s a quick summary of the equations you’ve derived:
- $v = u + at$
- $s = ut + \frac{1}{2}at^2$
- $v^2 = u^2 + 2as$
- As a consequence, however: $$v_{avg} = \frac{s}{t} = \frac{1}{t}\left(ut+\frac{1}{2}at^2\right) = u+\frac{1}{2}at = u+\frac{1}{2}(v-u) =\frac{v+u}{2} $$
- So, the average speed in uniformly accelerated motion over some interval is the average of initial and final speeds over that interval.
Remember, these equations only work when acceleration is constant.
Dimensional Analysis
- Dimensional analysis ensures the units on both sides of an equation are consistent.
- Let’s verify $s = ut + \frac{1}{2}at^2$:
- $s$ (displacement): $[\text{m}]$
- $ut$: $[\text{m s}^{-1}] \cdot [\text{s}] = [\text{m}]$
- $\frac{1}{2}at^2$: $[\text{m s}^{-2}] \cdot [\text{s}]^2 = [\text{m}]$
- Since all terms have the same unit $[\text{m}]$, the equation is dimensionally consistent.
- While dimensional analysis confirms consistency, it doesn’t guarantee correctness.
- Always check the logic behind the equation.
- However, if your equation is dimensionally inconsistent, i.e.:
- you are summing meters with seconds or apples with bananas and degrees Fahrenheit,
- or you have meters on RHS and pound sterlings on LHS,
- Your equations are truly wrong and there is definitely a mistake somewhere...
Limitations of the Kinematic Equations
The kinematic equations are powerful but rely on specific assumptions:
- Constant Acceleration: They don’t apply if acceleration varies.
- Straight-Line Motion: They’re valid only for one-dimensional motion.
- Neglect of External Forces: Air resistance and other forces are ignored.
Don’t use these equations for situations like circular motion or free fall with significant air resistance.
- In many real-life scenarios, air resistance is so small that it doesn’t significantly affect motion. For example:
- A ball dropped from a short height
- A car accelerating at low speeds
- A projectile moving slowly and close to the ground
- In these cases, the force of air drag is tiny compared to other forces like gravity, and neglecting it has little impact on the result.
- Air resistance becomes significant when:
- The object is moving very fast
- It has a large surface area
- It's moving through dense air
- But to describe this sort of motion mathematically, one needs more sophisticated approach using differential equations, which are out of scope of IB Physics syllabus.
- A car accelerates uniformly from $u = 10 \, \text{m s}^{-1}$ to $v = 30 \, \text{m s}^{-1}$ over a distance of $s = 80 \, \text{m}$. Find the acceleration.
- A ball is dropped from a height of $h = 20 \, \text{m}$. How long does it take to hit the ground? (Assume $g = 9.8 \, \text{m s}^{-2}$)
- A projectile is launched with an initial velocity of $u = 25 \, \text{m s}^{-1}$ at an angle of $30^\circ$ to the horizontal. Find the maximum height reached.
