Graphs of Motion — Position–Time and Velocity–Time

Imagine standing on a straight road and writing down where a car is every second. Plot those positions up the page and time across the page, join the dots, and you have a position–time graph. The clever part is that this single line records the whole trip — and just by looking at how it rises, falls, or stays flat, you can describe everything the car did without a stopwatch or a tape measure.

On an $x$–$t$ graph the vertical axis is position $x$ (measured along the line, using your sign convention) and the horizontal axis is time $t$. Three line shapes cover most rectilinear motion:

• Flat (horizontal) line → at rest. Position is not changing, so the object is parked.
• Straight sloping line → uniform (constant) velocity. Position changes by the same amount every second. A steeper line means a faster constant velocity.
• Curved line → changing velocity (accelerating). The steepness itself is changing, so the velocity is changing.

A line sloping up means the object is moving in the positive direction; a line sloping down means it is moving in the negative direction (back towards, or past, the start).

On a position–time ($x$–$t$) graph: a flat line = at rest; a straight sloping line = uniform (constant) velocity; a curved line = changing velocity (accelerating). Slope up = positive direction, slope down = negative direction. Steeper = faster.

Pause — copy the three $x$–$t$ line shapes (flat / straight slope / curve) and what each one means into your book before moving on.

We just saw that a steeper $x$–$t$ line means faster motion. That raises a question: can we turn "steeper" into an actual number — a velocity — instead of just "faster"? This card answers it → yes, the gradient (rise over run) of an $x$–$t$ graph is the velocity.

"Steeper means faster" is a feeling. Physics makes it a measurement. The steepness of a line is its gradient — how much it rises for each unit it runs across. On an $x$–$t$ graph the rise is a change in position and the run is a change in time, so the gradient is change-in-position over change-in-time. That is exactly the definition of velocity.

For a straight sloping line the gradient is the same everywhere, so the velocity is constant — read it off as $\Delta x / \Delta t$ between any two points. For a curved line the steepness changes, so the velocity changes; to find the velocity at one instant you draw a tangent at that point and take its gradient (the instantaneous velocity). A negative gradient (line sloping down) is a negative velocity — motion in the negative direction.

The gradient of a position–time graph is the velocity: $v = \dfrac{\Delta x}{\Delta t}$ (rise over run). Straight line → constant velocity (gradient the same everywhere); curve → changing velocity (use a tangent for the instantaneous value). A downward slope is a negative velocity.

Pause — copy the rule "gradient of $x$–$t$ = velocity" with the formula $v=\Delta x/\Delta t$ and the tangent point into your book before moving on.

We just saw that the gradient of an $x$–$t$ graph gives velocity. That raises a question: if we now plot that velocity against time, what does the gradient of this new graph tell us? This card answers it → the gradient of a velocity–time graph is the acceleration.

A velocity–time graph plots velocity $v$ up the page and time $t$ across. It answers a different question: not "where is it?" but "how fast, and is that changing?". The same gradient trick applies — but now the rise is a change in velocity and the run is a change in time, which is the definition of acceleration.

The line shapes mean something new on a $v$–$t$ graph:

• Flat line at zero → at rest (velocity is zero the whole time).
• Flat line above zero → uniform velocity (moving, but not speeding up — gradient is zero, so acceleration is zero).
• Straight sloping line → uniform (constant) acceleration. Upward slope = speeding up in the positive direction; downward slope = slowing down (or accelerating in the negative direction).

The gradient of a velocity–time graph is the acceleration: $a = \dfrac{\Delta v}{\Delta t}$. On a $v$–$t$ graph a flat line = constant velocity (zero acceleration); a straight sloping line = uniform acceleration (up = speeding up, down = slowing down).

Pause — copy "gradient of $v$–$t$ = acceleration", the formula $a=\Delta v/\Delta t$, and the flat/sloping $v$–$t$ shapes into your book before moving on.

We just saw that the gradient of a $v$–$t$ graph gives acceleration. That raises a question: the gradient uses the slope — but what does the area shut in beneath the line tell us? This card answers it → the area under a $v$–$t$ graph is the displacement.

Think about an object cruising at a constant 10 m/s for 4 seconds. Distance = speed × time = 10 × 4 = 40 m. On a $v$–$t$ graph that is a flat line at $v = 10$ for 4 seconds — and 10 × 4 is exactly the area of the rectangle under the line. That is no coincidence: the area under a velocity–time graph is always the displacement.

To find the area, split the region into rectangles and triangles:

• Rectangle (constant velocity): area $= v \times t$.
• Triangle (constant acceleration from or to rest): area $= \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} v t$.
• Trapezium / combined shapes: add up the rectangles and triangles that make up the region.

Area above the time axis is positive displacement; area below the axis (where velocity is negative) counts as negative displacement, because the object is moving the other way.

The area under a velocity–time graph is the displacement. Split it into rectangles (area $= vt$) and triangles (area $= \tfrac{1}{2}vt$) and add. Area above the time axis is positive; area below is negative (motion in the opposite direction).

Pause — copy the rule "area under $v$–$t$ = displacement", the rectangle and triangle area formulas, and the above/below-axis sign rule into your book before moving on.

We just saw that gradients and areas pull velocity, acceleration and displacement out of a graph. That raises a question: if these graphs all describe one motion, how do we redraw the same trip as the other kind of graph? This card answers it → read the gradient of the $x$–$t$ graph at each stage and plot it as the height of the $v$–$t$ graph.

An $x$–$t$ graph and a $v$–$t$ graph of the same trip look nothing alike — yet they hold the same story. The bridge between them is the gradient: the slope of the position graph becomes the height of the velocity graph. Read the slope of each section of the $x$–$t$ graph, and that number is how high to draw the $v$–$t$ line for that section.

The conversion rule, stage by stage

To convert $x$–$t$ → $v$–$t$, read the gradient of each section of the $x$–$t$ graph and plot it as the height of the $v$–$t$ line: flat → $v=0$; constant slope → flat $v$ line at that value; steepening curve → rising $v$ line. To go back, the area under the $v$–$t$ graph rebuilds the position change.

Pause — copy the conversion rule ("slope of $x$–$t$ becomes height of $v$–$t$") and the three-stage worked example into your book before moving on.