Uniformly accelerated motion and its graphs

Two cyclists currently have the same speed, but one is speeding up and the other is braking. A second later, their positions will differ. How can we describe what a single speedometer reading leaves out?

A journey may include stops and fast stretches. Total distance divided by total time gives the average speed, but does not tell us the velocity at each point. To find that, consider the average velocity over progressively shorter time intervals. Its limiting value as the interval approaches zero is the instantaneous velocity. A speedometer shows its magnitude, while the vector points along the tangent to the path. Equal average speeds do not mean identical journeys.

In 2 s, the velocity component rises from 3 to 7 m/s. If it changes uniformly, the increase each second is (7 − 3)/2 = 2 м/с. Acceleration describes this rate of change: aₓ = (vₓ − v₀ₓ)/t, in m/s². Twice the velocity change in the same time gives twice the acceleration. With constant acceleration, equal initial and final velocities mean zero acceleration. Without the uniform-change condition, the ratio gives only the average acceleration component.

You might think negative acceleration always slows an object down. Test that idea for an object moving in the negative direction: vₓ changes from −3 to −5 m/s. The component decreases, but the speed increases. When vₓ and aₓ have the same sign, the object speeds up. Opposite signs mean it slows down until a possible reversal. Acceleration describes a change in the velocity vector, so its sign must be read together with the direction of motion.

Suppose the velocity component increases by the same amount aₓ each second. Over t seconds the increase is aₓt, giving vₓ = v₀ₓ + aₓt. This describes uniformly accelerated motion: the acceleration vector is constant. Setting aₓ = 0 gives uniform motion; setting t = 0 gives the initial velocity. The units agree: м/с² · с = м/с. Doubling the time doubles the change in velocity, not necessarily the whole velocity.

Draw vₓ(t). The height of a point gives the current velocity component; the slope gives its change per unit time: aₓ = Δvₓ/Δt. Equal increases produce a straight line. A horizontal line at 5 m/s means motion at that velocity with zero acceleration. A steeper downward slope means more negative acceleration. Whether the object is speeding up or slowing down also depends on whether the line lies above or below zero.

How can a velocity graph give displacement? Over a short interval Δt, velocity is nearly constant, so its contribution is vₓΔt, the area of a narrow strip. Add the strips: the signed area under the graph gives the displacement component. Below the axis, vₓ is negative and the coordinate decreases. Distance requires the absolute values of the areas: earlier travel does not disappear when direction reverses. The area has units (м/с)·с = м.

With constant acceleration, the area under the straight vₓ(t) graph consists of a rectangle v₀ₓt and a signed triangular addition aₓt²/2. Hence sₓ = v₀ₓt + aₓt²/2. The factor one-half appears because the extra velocity grows from zero instead of acting in full from the start. Starting from rest, doubling the time makes displacement four times as large. If direction reverses, sₓ still gives the coordinate change, not the total distance travelled.

Sometimes time is not measured, but the initial and final velocities are known. Eliminating t from the two equations gives vₓ² − v₀ₓ² = 2aₓsₓ. For constant aₓ, this connects the change in squared velocity with acceleration and displacement. If aₓsₓ is negative, the squared velocity decreases. All components refer to the same axis. The final velocity squared does not by itself tell us the direction of motion.

There is another check: when velocity changes linearly, its average component is (v₀ₓ + vₓ)/2. Earlier values below the average balance later values above it. Multiply by t to obtain the same sₓ. For constant velocity, the average equals that velocity. After a reversal, the average component can be zero even though the distance travelled is positive; average speed must then be calculated separately.

A cyclist travels at 8 m/s and brakes with a constant acceleration of 2 m/s² opposite to the motion. Find the stopping time and distance.

Choose the forward direction as positive: v₀ₓ = 8 м/с, aₓ = −2 м/с². At rest, vₓ = 0. We must reduce the initial 8 m/s by 2 m/s each second.

From 0 = 8 − 2t, we get t = 4 с. The direction of motion does not change before this instant.

sₓ = 8 · 4 − 2 · 4²/2 = 16 м. This is also the distance travelled, since the cyclist does not reverse direction.

Check the area under the graph: the average speed is (8 + 0)/2 = 4 м/с, which gives 16 m in 4 s. We cannot extend the same acceleration beyond the stop without an additional condition.

The cyclist stops after 4 s, having travelled 16 m. During this time, speed decreases uniformly from 8 m/s to zero.

Acceleration describes a change in velocity. Its sign alone, without the velocity's sign, does not tell us whether speed is increasing.

The slope of a velocity graph gives acceleration; its signed area gives the change in coordinate.

Uniform-acceleration equations apply over an interval with constant acceleration. When motion reverses, distance and displacement must be calculated differently.

Next, apply velocity to falling, turning and changing the observer.