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Lesson 4 of 10

Motion in a Straight Line · Lesson 4 of 10

Acceleration and Motion Graphs

“Connect acceleration with velocity change, graph slopes, curvature and signed areas.”

Learning Objectives

• Calculate average and instantaneous acceleration. • Decide whether an object speeds up using both velocity and acceleration. • Connect position, velocity and acceleration graphs. • Obtain displacement from the signed area of a velocity-time graph.

Acceleration measures a change in velocity

Two cars can have the same velocity at one instant but behave differently immediately afterwards. One may maintain its velocity while the other speeds up. Acceleration describes this difference: it is the rate at which velocity changes with time. A change in velocity can involve speed, direction or both. In this chapter the motion lies along one fixed line, so direction is represented by a sign.

Average accelerationLaTeX
SI unit m s⁻²; dimension [L T⁻²]. An acceleration of +2 m/s² adds +2 m/s to velocity each second if it is constant.
Instantaneous accelerationLaTeX
Average acceleration is a secant slope on a velocity-time graph; instantaneous acceleration is the tangent slope.

The phrase “per second squared” does not mean that an object travels a square of time. Velocity already measures position change per second; acceleration measures how much that velocity changes per second. For example, velocities −6, −4 and −2 m/s at successive one-second instants show a constant +2 m/s² acceleration, although the object is initially moving in the negative direction.

Negative acceleration is not always slowing down

Velocity signAcceleration signWhat happens to speed
PositivePositiveSpeed increases
PositiveNegativeSpeed decreases until velocity reaches zero
NegativeNegativeSpeed increases
NegativePositiveSpeed decreases until velocity reaches zero

Away from v = 0, matching signs mean speed increases; opposite signs mean speed decreases. The sign of acceleration alone cannot answer that question. At v = 0, examine the motion on either side: with nonzero acceleration, an object may be turning around. “Retardation” or “deceleration” describes slowing down; when a numerical deceleration magnitude is quoted, it is normally positive even though the signed acceleration depends on the chosen axis.

Slowing down while acceleration is positive

Problem
Velocity changes uniformly from −12 m/s to −4 m/s in 2 s. What is acceleration and is the object speeding up?

  1. 1.The velocity change is −4 − (−12) = +8 m/s.
  2. 2.Acceleration is +8/2 = +4 m/s².
  3. 3.Speed falls from 12 to 4 m/s. Acceleration is positive but the object slows because its velocity is negative.

Reading all three graphs together

On a position-time graph, slope gives velocity. Curvature describes how that slope changes: concave upward means increasing velocity and positive acceleration; concave downward means decreasing velocity and negative acceleration. Do not confuse “curve rises” with “acceleration is positive”. A rising curve that gradually flattens has positive velocity but negative acceleration.

On a velocity-time graph, height gives velocity and slope gives acceleration. A horizontal line above zero is constant positive velocity, not rest. A sloping straight line means constant acceleration. Where the line crosses zero velocity, the object reverses if the sign changes. On an acceleration-time graph, height gives acceleration, and signed area gives change in velocity.

Areas and changesLaTeX
The first area has units (m/s) × s = m; the second has units (m/s²) × s = m/s.

You can understand the area rule without advanced calculus. Divide time into tiny intervals. In each interval, displacement is approximately velocity multiplied by the duration: a thin rectangle on the graph. Adding the rectangles and making them narrower approaches the exact signed area. Areas below the time axis are negative because the velocity there is negative.

Displacement and distance from the same graph

Distance from velocityLaTeX
Add the magnitudes of all velocity-time areas. Equivalently, find the area under the non-negative speed-time graph.
-6-30360123456Velocity (m/s)Time (s)
Constant negative acceleration through a reversal— The line crosses v = 0 at 3 s. The +9 m and −9 m triangular areas cancel for displacement, but add to 18 m distance.
An object returns to its start

Problem
For v = 6 − 2t m/s, calculate acceleration, displacement and distance from 0 to 6 s.

  1. 1.The slope of the velocity-time line is −2 m/s² throughout.
  2. 2.From 0 to 3 s, the triangular area is ½ × 3 × 6 = +9 m. From 3 to 6 s, the area is −9 m.
  3. 3.Displacement is 0 m and distance is 18 m. Average velocity is 0 m/s, while average speed is 3 m/s.
  4. 4.The object slows before 3 s and speeds up afterwards even though its acceleration stays negative. At 3 s, velocity is zero but acceleration is still −2 m/s².

What zero and constant quantities really imply

An object at rest throughout an interval has zero velocity and zero acceleration throughout that interval. Momentary rest does not imply zero acceleration. Similarly, zero average acceleration only means equal endpoint velocities; acceleration could have varied in between. Constant nonzero speed along one line with smooth motion implies constant direction and zero acceleration. A sudden idealised reversal would require a special collision or discontinuity; do not apply a differentiable-motion statement at that instant.

Quiz

Quick check

An object has v = −5 m/s and a = −2 m/s². What happens immediately afterwards?

Key Takeaways

Key Takeaways

• Acceleration is change in velocity per unit time. • Compare the signs of velocity and acceleration to decide whether speed rises or falls. • Slope of x–t gives v; slope of v–t gives a. • Signed area under v–t gives displacement; absolute area gives distance.