IB Physics A.1.2 Equations of Motion Notes

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.

Relating Velocity and Time

1. 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 $t$.
   1. How does its velocity change over time?
   2. What is the form of $v(t)$ ?
2. 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 that relates how velocity $v$ changes as a function of time $t$.

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 the previous section, we know that displacement is the area under a 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 a trapezium or, even simpler, of a rectangle and right-angle triangle. We obtain:

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

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

$$= ut + \frac{1}{2}at^2$$