The same motion can be described in several completely different ways: a motion diagram (dots showing position at equal time steps), a graph (position, velocity, or acceleration vs. time), an equation, or a plain-language narrative description. None of these is more "correct" than the others — they're translations of the same underlying physical situation.
For constant acceleration — and only constant acceleration — three equations describe instantaneous linear motion in one dimension. Each one connects a different trio of quantities, which is exactly why there are three: pick the one that already contains the variables you know and the one you're solving for.
A car starts at x₀ = 0 m with vₓ₀ = 5 m/s and accelerates at 2 m/s². Find its position after 4 seconds.
Near Earth's surface, every object in free fall — regardless of mass — experiences a constant downward acceleration due to gravity.
Free fall is just the constant-acceleration kinematic equations from the previous section, with a = −g substituted in (using whichever sign convention you've chosen for "up"). Nothing new to memorize — only a specific number to plug in.
Graphing position, velocity, and acceleration as functions of time reveals the relationships between them directly — without needing to be handed an equation at all. The next two sections show exactly how: slopes move you down the chain (position → velocity → acceleration), and areas move you back up (acceleration → velocity → position).
An object's instantaneous velocity is the slope of the line tangent to a point on a position-vs-time graph:
An object's instantaneous acceleration is the slope of the line tangent to a point on a velocity-vs-time graph:
Use the tool below to see both at once. Drag the time slider on a thrown-ball scenario and watch the tangent line on each graph — its slope is exactly the value plotted on the graph beneath it.
A ball is thrown straight up at 20 m/s (a = −10 m/s², using g ≈ 10 m/s²). Drag the time slider — the amber tangent line on each graph shows the slope at that instant, which equals the value on the graph below it.
Run the slope relationship in reverse and you get areas. The displacement of an object during a time interval equals the area under a velocity-vs-time graph over that interval:
Likewise, the change in velocity during a time interval equals the area under an acceleration-vs-time graph over that interval:
Try it below: adjust the boundaries on the same thrown-ball's velocity graph and compare the shaded area to the exact displacement from the kinematic equations.
Same thrown ball, v(t) = 20 − 10t. Set the boundaries t₁ and t₂ — the shaded area (signed: below the axis counts negative) equals the displacement over that interval.
The two numbers match — the area under a velocity-time graph really is the displacement, whether you compute it as an integral or read it straight off the graph.
An object's velocity is given by v(t) = 4t (m/s) for 0 ≤ t ≤ 3 s. Find its displacement over this interval using the area under the graph.