The gradient of a displacement-time graph gives the velocity
Motion & Kinematics - OCR A-Level Physics
Key rule
The gradientThe rate of change of one variable with respect to another; the slope of a graph at a point. of a displacement-time graph at any point equals the velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. at that point. Steeper gradient $\Rightarrow$ faster motion. Negative gradient $\Rightarrow$ motion in the opposite direction.
$$v = \frac{\Delta s}{\Delta t}$$
Diagram pending
Four-panel displacement-time graph. Panel 1: straight line through origin (constant velocity). Panel 2: horizontal line (stationary). Panel 3: upward curve (accelerating). Panel 4: line with negative gradient (returning to start). Tangent triangle drawn on the curved panel for instantaneous velocity.
Will be replaced with a GeoGebra SVG in stream 2.
Case 1: Straight line through origin
Constant velocity. The slope is the velocity, given by $v = \frac{s_2 - s_1}{t_2 - t_1}$. Example: line from $(0, 0)$ to $(10, 50)$ gives $v = 5 \text{ m s}^{-1}$.
Case 2: Horizontal line
Zero gradient means zero velocity. The object is stationaryNot moving; the object has zero velocity.. Time is passing but displacement is unchanged.
Case 3: Curve
Changing gradient means changing velocity, so the object is accelerating or decelerating. An upward-curving line (increasing gradient) means acceleration. A line that bends over (decreasing gradient) means deceleration. For instantaneous velocityThe velocity at a specific moment in time, given by the gradient of a displacement-time graph at that point., draw a tangent at that time and find its gradient.
Case 4: Negative gradient
Velocity in the opposite direction to the chosen positive convention. The object is moving back toward (or past) the origin. The magnitude of the gradient is still the speed.
- An object moving at $4 \text{ m s}^{-1}$ for $3 \text{ s}$, then stationary for $2 \text{ s}$, then returning at $-6 \text{ m s}^{-1}$ for $2 \text{ s}$: the s-t graph has three straight segments with gradients $+4$, $0$, and $-6$.
- To find instantaneous velocity at $t = 5 \text{ s}$ on a curve: draw a tangent that just touches the curve at $t = 5 \text{ s}$, then $v = \frac{\Delta s}{\Delta t}$ from a large triangle on that tangent.
- Average velocity over an interval is the gradient of the chord (straight line) between the two endpoints, not the tangent.
Common Mistake
MEDIUM
Wrong: Reading the value of $s$ off the graph at a point and calling it the velocity. Or drawing a tiny gradient triangle that gives low accuracy.
Right: Velocity is the gradient (slope), not the height. Use a triangle spanning at least half the graph for accuracy: $v = \frac{\Delta s}{\Delta t}$.
Right: Velocity is the gradient (slope), not the height. Use a triangle spanning at least half the graph for accuracy: $v = \frac{\Delta s}{\Delta t}$.
Examiner Tips and Tricks
- When asked for instantaneous velocity from a curved s-t graph, draw the tangent at the specified time and show a large gradient triangle clearly on the graph.
- OCR awards method marks for a correctly drawn tangent even if the read-off is slightly out.
- State units of $\text{m s}^{-1}$ on every velocity answer.