Motion, speed and acceleration

Motion, speed and acceleration

Mechanics begins with a careful description of motion. Distance is the total length of the route travelled and is a scalar. Displacement is the change in position from start to finish and includes direction, so it is a vector. A student who walks 20 metres east and then 20 metres west travels 40 metres, but returns to the starting point and has zero displacement.

Average speed is total distance divided by total time. Average velocity is displacement divided by elapsed time. Use compatible units before calculating: kilometres per hour can be converted to metres per second by multiplying by 1,000/3,600. Instantaneous speed refers to a particular moment. Do not average two speeds directly unless the time spent at each speed is the same.

Acceleration is the rate of change of velocity. In one dimension, where u is the initial velocity and v is the final velocity. An acceleration can be negative relative to a chosen positive direction. Negative acceleration does not always mean slowing down; slowing depends on whether velocity and acceleration point in opposite directions.

For constant acceleration, useful equations include, and. They share the assumption that acceleration remains uniform during the interval. Select the equation containing the known quantities and the required unknown. State the positive direction, particularly in vertical motion, before substituting signed velocities and gravitational acceleration.

On a displacement–time graph, the gradient gives velocity. On a velocity–time graph, the gradient gives acceleration and the signed area under the graph gives displacement. A horizontal velocity–time line means constant velocity, rather than zero motion. Areas below the time axis contribute negative displacement; the total distance requires adding the magnitudes of the separate positive and negative areas.

Every numerical answer needs a unit and a sensible interpretation. A car increasing uniformly from rest to 12 m/s in 4 s accelerates at 3 m/s² and covers 24 m during that time. The calculation uses its average velocity of 6 m/s. Sketching a velocity–time triangle gives the same distance and provides an independent check on the algebra.

Choosing a frame and keeping signs consistent

Begin a motion problem by defining the object, interval and positive direction. The origin is a chosen reference point, not necessarily where the object began moving. If east is positive, a westward velocity is negative even when the object speeds up. A body with u = −4 m/s and reaches v = −10 m/s after three seconds: its speed increases. Conversely, u = −10 m/s with a = +2 m/s² describes slowing until it stops. Write signs with the values rather than trying to attach a direction after an unsigned calculation. A sketch of the positions often resolves an apparent contradiction.

Deriving the constant-acceleration relationships

Uniform acceleration makes the velocity–time graph a straight line. Its gradient gives, leading directly to. The area under that straight line is a trapezium, so displacement is. Substitution of gives s = ut + ½at². Eliminating time between the first two relationships gives. These are not four unrelated rules to memorise. They are connected descriptions of one model. If resistance changes acceleration substantially, use a graph or a more suitable model rather than applying them automatically.

Reading a journey in stages

A vehicle can accelerate, cruise and brake within one journey. Treat each interval separately, using the final velocity of one as the initial velocity of the next. A velocity–time graph with a triangle, rectangle and triangle represents three separate displacements whose signed areas are added. On a curved displacement–time graph, average velocity is the gradient of a chord joining two points, while instantaneous velocity is the gradient of a tangent at one point. A steeper negative gradient means a larger speed in the negative direction. Do not interpret the height of a displacement graph as speed or confuse a flat displacement graph with a flat velocity graph.

Free fall and vertical projection

Near Earth's surface, neglecting air resistance, all freely falling bodies have the same downward acceleration g. A heavier object does not have a larger gravitational acceleration merely because its weight is greater. For upward-positive motion use throughout both ascent and descent. At the highest point the vertical velocity is momentarily zero, but acceleration remains downward. A ball thrown upward at 20 m/s with takes two seconds to reach its highest point and rises 20 m. If it returns to the launch level, its ideal total flight time is four seconds and its final velocity is −20 m/s.

Projectile motion as two linked components

For a projectile under gravity alone, horizontal acceleration is zero while vertical acceleration is −g. Resolve launch velocity into u cos θ horizontally and u sin θ vertically. The same time t belongs to both components: and. If launch and landing are at the same level, symmetry gives flight time 2u sin θ/g. The familiar range formula u² sin 2θ/g requires that level-ground assumption. From a cliff or onto a higher platform, first solve the vertical equation for the appropriate positive time, then calculate horizontal distance.

Average speed and practical measurements

A runner covering equal distances at two different speeds does not spend equal times at those speeds. Calculate total distance and total time explicitly. A 100 m outward trip at 5 m/s takes 20 s, and a 100 m return at 2 m/s takes 50 s. Average speed is 200/70 = 2.86 m/s, whereas average velocity is zero because final displacement is zero. Experimental speed can be measured using a measured track and stopwatch or a light gate of known interrupted length. Repeat measurements and state timing uncertainty. Short manual timings have large percentage errors because reaction time becomes a large fraction of the interval.

Checking an exam solution

Identify whether the question asks distance, displacement, speed, velocity or acceleration before selecting a formula. A negative solution for time often represents an earlier mathematical intersection outside the required interval and should be interpreted, not reported blindly. A value of v² cannot be negative in a real-motion result; such a value can mean that the assumed position cannot be reached. Convert 72 km/h to 20 m/s before substitution. Keep extra digits during intermediate steps and round the final answer sensibly. If a graph is provided, quote its scales and use a large gradient triangle instead of two adjacent points.

Self-check and explanation

Suppose a trolley moves at 3 m/s for four seconds and then accelerates uniformly to 7 m/s over two seconds. The first displacement is 3 × 4 = 12 m. The second is ½(3 + 7) × 2 = 10 m, giving 22 m in six seconds and average velocity 3.67 m/s in the original direction. The acceleration in the second stage is 2 m/s². A separate question asks whether a stone at its highest point has zero force. It does not: its weight still acts downward. Zero instantaneous velocity describes the motion at that instant; zero resultant force would describe its acceleration.