A.1.2 Equations of motion

The Derivation and Application of Kinematic Equations

1. You're standing at a traffic light, watching a car accelerate smoothly from rest.
2. 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

1. Imagine 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 $t$.
2. How does its velocity change over time? What is the form of $v(t)$ ?
3. Acceleration is defined as the rate of change of velocity:

$$a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}⟹a=\frac{v(t)-u}{t}⟹v(t)=u+at$$

This is the equation, which relates states how velocity $v$ changes as a function of time $t$.

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.

1. From previous section, we know, that displacement is actually the area under velocity-time graph.
2. We have also established the velocity-time dependence as a linear function $v(t)$.
3. 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:

$$\text{Area}=s(t)=ut+\frac{1}{2}(v(t)-u)t$$

$$=ut+\frac{1}{2}(u+at-u)t$$

$$=ut+\frac{1}{2}a{t}^2$$

This is the equation, which relates displacement to time in uniformly accelerated motion.

3. Relating Velocity and Displacement

Finally, let’s connect velocity and displacement without involving time.

1. We start with two standard equations: $$v=u+at$$ $$s=ut+\frac{1}{2}a{t}^2$$
2. Our goal is to eliminate time $t$.
3. We first express $t$ from first equation: $$t=\frac{v-u}{a}$$
4. Replace $t$ in the displacement formula: $$s=u\left(\frac{v-u}{a}\right)+\frac{1}{2}a{\left(\frac{v-u}{a}\right)}^2$$
5. Now, we simplify each term: $$s=\frac{u(v-u)}{a}+\frac{(v-u{)}^2}{2a}$$
6. And combine over the common denominator: $$s=\frac{2u(v-u)+(v-u{)}^2}{2a}$$
7. Factor the numerator:$$s=\frac{(v-u)[2u+(v-u)]}{2a}=\frac{(v-u)(v+u)}{2a}$$
8. Now use the identity $(v-u)(v+u)=v^2-u^2$:
   $$s=\frac{v^2-u^2}{2a}$$
9. Finally, rearrange: $$v^2=u^2+2as$$

Summary of Kinematic Equations

1. Here’s a quick summary of the equations you’ve derived:
   1. $v=u+at$
   2. $s=ut+\frac{1}{2}a{t}^2$
   3. $v^2=u^2+2as$
2. As a consequence, however: $$v_{avg}=\frac{s}{t}=\frac{1}{t}(ut+\frac{1}{2}a{t}^2)=u+\frac{1}{2}at=u+\frac{1}{2}(v-u)=\frac{v+u}{2}$$
3. So, the average speed in uniformly accelerated motion over some interval is the average of initial and final speeds over that interval.

Dimensional Analysis

1. Dimensional analysis ensures the units on both sides of an equation are consistent.
2. Let’s verify $s=ut+\frac{1}{2}a{t}^2$:
   1. $s$ (displacement): $[\text{m}]$
   2. $ut$: $[\text{m/s}]\cdot [\text{s}]=[\text{m}]$
   3. $\frac{1}{2}a{t}^2$: $[{\text{m/s}}^2]\cdot [\text{s}{]}^2=[\text{m}]$
3. Since all terms have the same unit $[\text{m}]$, the equation is dimensionally consistent.

Limitations of the Kinematic Equations

The kinematic equations are powerful but rely on specific assumptions:

1. Constant Acceleration: They don’t apply if acceleration varies.
2. Straight-Line Motion: They’re valid only for one-dimensional motion.
3. Neglect of External Forces: Air resistance and other forces are ignored.