Cambridge Syllabus Notes

Equations of Motion: Finding Displacement from a Velocity–Time Graph 🚀

Exam Tip: Remember that the area under a velocity–time graph always gives you the displacement (not distance). Keep track of the sign of the area: positive area → forward, negative area → backward.

1️⃣ What is a Velocity–Time Graph?

A velocity–time graph shows how an object's speed (and direction) changes over time. The vertical axis is velocity v (m/s) and the horizontal axis is time t (s). Think of it like a road map where the height of the line tells you how fast you’re going at each moment.

2️⃣ The Area Under the Curve = Displacement 📈

The key rule: $$\text{Displacement } s = \int v(t)\,dt$$ In a graph, this integral is simply the area between the curve and the time axis. • If the curve lies above the axis, the area is positive (forward motion). • If it lies below, the area is negative (backward motion). Analogy: Imagine a water tank. The area is the amount of water that flows in (positive) or out (negative) over time.

3️⃣ Step‑by‑Step: Calculating Displacement

Identify the segments of the graph where the velocity is a simple shape (rectangles, triangles, trapezoids).
Calculate the area of each segment using the appropriate formula:
- Rectangle: A = width × height
- Triangle: A = ½ × base × height
- Trapezoid: A = ½ × (top + bottom) × height
Assign a positive sign to areas above the axis and a negative sign to areas below.
Sum all signed areas to get the total displacement.

4️⃣ Example Problem

A car starts from rest, accelerates at a constant rate of 2 m/s² for 5 s, then travels at a constant speed of 10 m/s for 8 s, and finally stops over 4 s. Sketch the velocity–time graph and find the total displacement.

Segment	Description	Area (m)
1	Acceleration from 0 to 10 m/s (triangle)	½ × 5 s × 10 m/s = 25 m
2	Constant speed 10 m/s for 8 s (rectangle)	10 m/s × 8 s = 80 m
3	Deceleration from 10 m/s to 0 (triangle, negative)	-½ × 4 s × 10 m/s = -20 m
Total Displacement		25 m + 80 m – 20 m = 85 m

The car travels a total of 85 m forward. Notice how the negative area from the deceleration segment reduces the total.

5️⃣ Common Pitfalls & Exam Tips 🧠

Mixing distance and displacement: Always check the sign of the area.
Ignoring units: Velocity is in m/s, time in s, so area (m/s × s) gives metres.
Overlooking small segments: Even a brief negative area can change the final answer.
Sketching first: A quick sketch helps identify shapes and signs.
Check your arithmetic: A single mis‑calculated area can throw off the whole answer.

Quick Review: • Area = displacement • Positive area → forward, negative area → backward • Break the graph into simple shapes, calculate each area, sum with signs • Keep units consistent and double‑check signs

determine displacement from the area under a velocity–time graph