Displacement–Time and Velocity–Time Graphs: MDCAT Physics notes
Displacement–Time and Velocity–Time Graphs for MDCAT: slope as velocity and acceleration, area under the v–t graph, and distance–time slopes.
Reading any motion graph
Two quantities can be read from a motion graph: the slope (gradient), which is rise/run, and the area under the line. What they mean depends on the axes.
| Graph | Slope gives | Area gives |
|---|---|---|
| Displacement–time | Velocity | No useful meaning |
| Distance–time | Speed | No useful meaning |
| Velocity–time | Acceleration | Displacement (distance covered when there is no reversal) |
Displacement–time graphs
- Horizontal line: slope zero, body at rest.
- Straight inclined line: constant slope, so constant (uniform) velocity. A line at 45$^\circ$ has slope $\tan45^\circ = 1$ (in the units of the axes): the velocity is constant, neither increasing nor decreasing.
- Steeper line: greater velocity.
- Curve bending upward: slope increasing, so the body is accelerating.
- Curve flattening: slope decreasing, so the body is slowing down.
- Line sloping downward: negative velocity, i.e. moving back towards the origin.
Distance–time graphs
Distance never decreases, so a distance–time graph can rise or stay flat but can never slope downward. Its slope (speed) is positive while the body moves and zero when it is at rest, but never negative. A displacement–time graph, by contrast, can have a negative slope.
Velocity–time graphs
- Horizontal line: constant velocity, zero acceleration.
- Straight inclined line: constant slope, so constant (uniform) acceleration. A straight v–t line tells you nothing more; it does not mean the body never turned round or had zero displacement.
- Downward straight line: uniform deceleration.
- Curve: slope changes, so acceleration is variable.
- Area under the graph = velocity × time = displacement; for motion in one direction this equals the distance covered.
Worked examples
1. A displacement–time graph is a straight line from (0 s, 0 m) to (5 s, 40 m). Slope $= 40/5 = 8\ \mathrm{m\,s^{-1}}$, constant velocity.
2. A velocity–time graph rises in a straight line from 0 to 12 m/s in 4 s, then stays at 12 m/s for 6 s.
- Acceleration in the first part = slope $= 12/4 = 3\ \mathrm{m\,s^{-2}}$; in the second part it is 0.
- Distance in the first part = triangle area $= \tfrac12\times4\times12 = 24$ m.
- Distance in the second part = rectangle area $= 6\times12 = 72$ m.
- Total distance = 96 m.
Key formulas
- Slope of $d$–$t$ graph: $v = \dfrac{\Delta d}{\Delta t}$
- Slope of $v$–$t$ graph: $a = \dfrac{\Delta v}{\Delta t}$
- Area of $v$–$t$ graph: $S = $ area of triangles + rectangles (+ trapezia, $\tfrac12(v_i+v_f)t$)
Common MDCAT traps
- Constant slope on a displacement–time graph means constant velocity, not constant acceleration.
- Constant slope on a velocity–time graph means constant acceleration; the v–t graph for constant acceleration is a straight line, not a parabola. (It is the displacement–time graph that becomes a parabola.)
- Area under a v–t graph is distance/displacement, not speed or acceleration.
- The slope of a distance–time graph can be zero but never negative.
Quick revision
- $d$–$t$ slope = velocity.
- $v$–$t$ slope = acceleration.
- $v$–$t$ area = displacement.
- Straight $v$–$t$ line = uniform acceleration; curve = variable acceleration.
- Horizontal $d$–$t$ line = rest.