Distance is the total length of the path you travel. It does not include direction. Displacement is the change from your starting position to your finishing position, including direction. On a straight east–west line, you can calculate displacement by treating east as positive and west as negative. For example, walking 70 m east and 20 m west gives a distance of 90 m, but a displacement of 50 m east.

Average speed tells you how much distance is travelled per second over the whole journey. Average velocity tells you how quickly your position changes and in which direction. Both use the entire journey time, including stops. A person may travel a considerable distance but have zero displacement if they finish at their starting point. You can rearrange average speed = distance ÷ time to find time = distance ÷ average speed.

Uniform motion along a straight line means that an object travels equal distances in equal time intervals without changing direction. Many real journeys are not uniform: a bus moves between stops, waits for passengers and changes speed in traffic. A speed calculated for only one section is not automatically the same as the average speed for the whole journey.

To find average speed, first add every distance travelled. Next add the complete journey time, including waiting or stopping. Divide the total distance by the total time. Be careful with units: 1 km = 1000 m and 1 min = 60 s. Simply averaging several speed values can give an incorrect result when the sections have different durations. A stop adds time, but it adds no distance. Example (not your problem): 600 m in 2 min, then a 1 min wait, gives 600 ÷ 180 ≈ 3.3 m/s. On a speed–time graph, time is on the horizontal axis and speed on the vertical axis; a stop is a flat line at zero speed, and the area under the line is the distance travelled.

A position–time graph shows an object’s position on the vertical axis and the time on the horizontal axis. For motion at constant velocity, the graph is a straight line. Its gradient tells you the velocity: divide the change in position by the change in time. A line rising steeply represents a greater positive velocity than a gently rising line. A horizontal line represents a stationary object.

The position where a line starts tells you the object’s initial position, not how quickly it moves. If two position–time lines cross, the objects share the same position at the same time. This is when they meet or one catches the other, provided their positions are measured from the same origin. You can extend straight lines to predict what happens later, but that prediction assumes the velocities stay constant. Example (not your problem): a line from (0 s, 5 m) to (4 s, 13 m) has gradient (13 − 5) ÷ 4 = 2 m/s. To find where two lines meet, extend a table of positions until both rows show the same value.

A velocity–time graph has time along the horizontal axis and velocity along the vertical axis. A horizontal section shows constant velocity. A line that slopes upward shows increasing velocity; a line that slopes downward shows decreasing velocity. For a straight-line section, the gradient gives acceleration: change in velocity divided by change in time. The acceleration may be positive, zero or negative.

The area between the line and the time axis represents displacement. When the velocity is positive for the entire journey, this area is also the distance travelled. You can calculate areas by dividing a graph into rectangles and triangles. Rectangle area = width × height; triangle area = ½ × base × height. Always label your axes and use seconds and metres per second, so an area is measured in metres. Example (not your problem): 0 to 6 m/s in 3 s, then 6 m/s for 10 s: triangle ½ × 3 × 6 = 9 m, rectangle 10 × 6 = 60 m, total 69 m. The rectangle’s width is its own duration (10 s), not the whole journey time.

Velocity is the rate of change of position, with direction (average velocity = displacement ÷ time, unit m/s). Acceleration describes how quickly velocity changes over time. An object moving quickly does not necessarily have a large acceleration: it might be travelling at a constant velocity. Likewise, a slowly moving object can accelerate rapidly. If an object’s velocity increases by the same amount every second, its acceleration is constant. Acceleration is measured in metres per second squared (m/s²).

One way to recognise changing motion is to record positions after equal time intervals. A cart travelling at a constant speed in one direction makes equally spaced marks. If the gaps between marks become larger, the distance travelled during each interval is increasing. The position–time graph will curve upward for a uniformly accelerating object moving in the positive direction; a velocity–time graph for constant acceleration is a straight sloping line. Describing this pattern does not require explaining what causes the motion. Example (not your problem): positions 0, 1, 3, 6 m at 0, 1, 2, 3 s give intervals of 1, 2, 3 m. Each interval is 1 m longer, so the object is speeding up. Plot the points as given; do not force a straight line through them.

A cart starts from rest and its velocity increases by the same amount every second. This is constant (uniform) acceleration. Its velocity–time graph is a straight line starting at zero. The velocity at the end, v, is the highest velocity the cart reaches.

If the cart keeps moving in one direction, the area under the graph is the distance travelled. The area is a triangle: ½ × t × v. A cart moving at its final velocity for the whole time would travel t × v, so the triangle is half of that rectangle. Average velocity = distance ÷ time = ½ × v. So the average velocity is half the final velocity, and the final velocity is double the average velocity.

Important: this shortcut works only if the cart starts from rest, has constant acceleration and does not reverse. If it starts with velocity u instead, average velocity = (u + v) ÷ 2. Example (not your problem): from rest to 8 m/s in 5 s gives average velocity 4 m/s and distance 4 × 5 = 20 m.