Connect acceleration, velocity and position

A negative acceleration need not mean slowing down

An object starts at position zero with velocity +2 m/s. Its acceleration is −1 m/s² for four seconds. Predict its velocity after each second. The values are +2, +1, 0, −1 and −2 m/s. During the first two seconds its speed decreases; during the last two its speed increases in the negative direction. Acceleration stayed negative throughout.

Velocity describes position change per unit time. Acceleration describes velocity change per unit time. The unit m/s² means that each second changes velocity by some number of metres per second. It does not mean position changes by the same amount each second. The OpenStax constant-acceleration section provides the standard equations; this widget uses an original idealized trajectory and numerical cases.

Build velocity from repeated equal changes

Let v₀ mean velocity at time zero, a constant acceleration, t elapsed time, and v(t) velocity at that time. Then v(t) = v₀ + at. Read this as “initial velocity plus the accumulated change in velocity.” The product at has units (m/s²) × s = m/s, matching the velocity units.

For v₀ = +2 m/s and a = −1 m/s², at t = 3 s the result is 2 + (−1) × 3 = −1 m/s. This is not a speed of negative one: speed is the magnitude, 1 m/s. The negative sign identifies direction along the chosen axis.

Your action: keep initial velocity at +2 m/s and compare acceleration −1, 0 and +1 m/s². The velocity graphs are straight lines because equal time intervals produce equal velocity changes. Zero acceleration gives constant velocity; it does not require the object to be at rest. Positive acceleration with positive velocity increases speed. Negative velocity with positive acceleration initially decreases speed until velocity reaches zero, after which the same acceleration can increase speed in the positive direction.

Position comes from the changing velocity

At constant acceleration, velocity changes linearly in time. Its average over an interval starting at zero is (v₀ + v(t))/2. Displacement equals that average velocity times elapsed time. With initial position zero, substitution gives:

x(t) = v₀t + ½at²

Here x(t) is position in metres. The first term is the position change that the initial velocity alone would produce; the second accounts for the continuing velocity change. Both terms have units of metres. This equation assumes a is constant throughout the interval. An average acceleration by itself does not generally justify using it when the acceleration varies.

For the default case at t = 3 s, x = 2 × 3 + ½ × (−1) × 3² = 6 − 4.5 = 1.5 m. At t = 4 s, x = 8 − 8 = 0 m. The object has returned to the starting position, despite moving throughout most of the interval. The exact integer-second position sequence is 0, 1.5, 2, 1.5, 0 m.

Connect both graphs without confusing their slopes

The slope of a position–time graph is velocity. At first the default position curve rises, then flattens at its maximum, then falls. The slope of a velocity–time graph is acceleration. The velocity graph keeps a constant downward slope even where it crosses zero.

The position and velocity plots use different vertical units and fixed ranges. Their slopes cannot be compared by screen angle alone. Read the axis labels and calculate change divided by elapsed time. In a one-second interval from 0 to 1 s, position changes by 1.5 m, giving an interval-average velocity of 1.5 m/s. The instantaneous velocities at the endpoints are 2 and 1 m/s. An interval slope need not equal either endpoint value.

A signed area under the velocity graph gives displacement. From 0 to 2 s, the triangle above zero has area ½ × 2 s × 2 m/s = +2 m. From 2 to 4 s, the triangle below zero contributes −2 m. The signed sum is zero displacement. Adding the area magnitudes instead gives 4 m of travel. Area units are seconds times metres per second, or metres.

Separate the exact model from its drawing

The curve is evaluated at closely spaced display times, but those plotted segments are only a rendering. Positions, velocities and the reversal time come from analytic equations. The model does not infer acceleration from noisy measurements or approximate the trajectory by repeatedly updating position with a finite time step. A real experiment would need instrument calibration, timing evidence and uncertainty analysis.

The widget prescribes acceleration without explaining what produces it. No friction, motor, spring, gravity field or collision is simulated. Later force models must supply that cause and justify when the constant-acceleration approximation is adequate.

Transfer: An object starts at −3 m/s with acceleration +0.5 m/s² and initial position zero. Find velocity and displacement after four seconds. Is it speeding up or slowing down over that interval? State whether it has reversed yet.

Check the acceleration transfer

Velocity is −3+0.5×4 = −1 m/s. Displacement is −3×4+½×0.5×16 = −8 m. The object is still moving negatively, but its speed has decreased from 3 to 1 m/s. It would reach zero velocity at 6 s if the model continued, so it has not reversed within four seconds.