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.