Uniformly accelerated motion
Three equations, v = v₀ + at, x = x₀ + v₀t + ½at² and v² = v₀² + 2aΔx, look like three formulas to learn by heart. They are really one: the acceleration is constant, and everything follows from that.
0Position, velocity, acceleration
Kinematics describes motion without asking what causes it. Three quantities are enough, each one being the rate of change of the one before.
| Quantity | Meaning and unit |
|---|---|
| x(t) | The position along a directed axis, in metres (m), at time t in seconds (s). |
| v = dx/dt | The velocity: how fast the position changes, in metres per second (m/s). It carries a sign. |
| a = dv/dt | The acceleration: how fast the velocity changes, in metres per second squared (m/s²). |
When the acceleration is zero, the velocity does not change and the motion is uniform motion in a straight line: x = x₀ + vt. When it is constant but not zero, the motion is said to be uniformly accelerated. Steady braking and a stone dropped in a vacuum are both examples.
1Where the equations come from
All three formulas follow from a single assumption: a is constant.
If a is constant, v changes by the same amount every second: v(t) = v₀ + at, where v₀ is the velocity at t = 0. The graph of v against t is a straight line of gradient a. The position follows, since x is the antiderivative of v: x(t) = x₀ + v₀t + ½at².
The factor ½ comes from an area. Between 0 and t, the displacement is the area of the trapezium under the line, that is, the average velocity (v₀ + v)/2 multiplied by t, which gives v₀t + ½at². This is the link between derivative and antiderivative seen in derivatives and antiderivatives.
The third equation gets rid of time. From v = v₀ + at we get t = (v − v₀)/a. Substituting into Δx = (v₀ + v)/2 × t gives Δx = (v² − v₀²)/2a, in other words v² = v₀² + 2aΔx. It is the one to reach for when the problem gives a distance and never mentions a duration.
2Free fall and vertical launch
Free fall is the textbook case: with no air resistance, every body falls with the same acceleration, written g.
Near the surface of the Earth, g is about 9.81 m/s² (the conventional standard value is 9.80665 m/s²). Mass plays no part: a marble and an anvil released in a vacuum land together.
A stone is dropped from 20 m with no initial velocity. With the axis pointing downwards, x = ½gt², so t = √(2 × 20 / 9.81) ≈ 2.02 s. The landing speed is v = gt ≈ 19.8 m/s, roughly 71 km/h. The third equation confirms it without using time: v = √(2gh) = √392.4 ≈ 19.8 m/s.
3Reading a v(t) graph
Two readings are all you need on a graph of velocity against time.
The gradient of the curve is the acceleration. The area between the curve and the time axis is the displacement, counted as negative below the axis: displacement and distance travelled can therefore differ.
A car goes from 0 to 20 m/s in 8 s. The gradient is 20/8 = 2.5 m/s². The area of the triangle is ½ × 8 × 20 = 80 m, the same result as ½at² = ½ × 2.5 × 64. For braking from 25 m/s (90 km/h) with a = −5 m/s², the vehicle stops in 5 s, after 25² / (2 × 5) = 62.5 m, not counting the driver's reaction time.
4The lab
Enter v₀, a and t: the graphs and the final values are computed, and the third equation is checked live.
5The traps
6Test yourself
Five questions, one correct answer each time.
7Glossary
| Term | Meaning |
|---|---|
| kinematics | the study of motion without regard to its causes |
| uniformly accelerated rectilinear motion | straight-line motion with constant acceleration |
| velocity | the derivative of position with respect to time, in m/s |
| acceleration | the derivative of velocity with respect to time, in m/s² |
| displacement | the change in position x − x₀, with its sign |
| free fall | fall under gravity alone, with no friction |