Two golden rules, and nearly every mark on this topic depends on knowing them cold:

• Gradient = acceleration (units: $\text{m/s}^2$)
• Area under the graph = distance travelled

Break the graph into rectangles (constant velocity), triangles (acceleration from/to rest), and trapeziums (acceleration between two non-zero velocities), then sum their areas.

A faster approach: the trapezium formula $\frac12(v_{\text{initial}}+v_{\text{final}})\times t$ actually works for all three shapes (a rectangle is just a trapezium with $v_1=v_2$; a triangle is one with $v_1=0$), so you can apply one formula to every straight segment.

Worked example: A journey accelerates 0–10s from 0 to 20 m/s, holds constant 10–30s, then decelerates 30–40s to 10 m/s.

Total distance = 650 m.

A curve means the gradient — and so the acceleration — is continuously changing. You can no longer state "the" acceleration over an interval; you can only estimate the instantaneous acceleration at one specific moment.

Draw a tangent at the required point, form a large gradient triangle on it, and calculate rise/run. A large triangle minimises reading error.

Split the time axis into equal-width strips (the question tells you how many), read the height (velocity) at each boundary, then sum the trapezium area of each strip:

Worked example: Using 5 strips over 10 seconds, strip width $h=2$s. Read velocities $v_0,v_2,v_4,v_6,v_8,v_{10}$ from the graph, then sum $\frac12(v_0+v_2)(2)+\frac12(v_2+v_4)(2)+\dots$ for all five strips.

Over- or under-estimate? If the curve bends down (concave down), the trapezium tops sit under the curve → underestimate. If it bends up (concave up), the tops sit above the curve → overestimate.

The tangent/trapezium methods are physical approximations of differentiation and integration:

Given $v=t^3-4t^2+2t$, the exact acceleration at $t=5$ comes from differentiating: $a=3t^2-8t+2$, giving $a=3(25)-40+2=37\ \text{m/s}^2$ — no tangent-drawing required. The graphical method is simply an estimate of what differentiation gives exactly.

A sprinter runs 200 m. The velocity-time graph is a triangle (base 5s) followed by a rectangle (base 15.32s), both reaching the same maximum speed $x$. Find $x$.