Kinematics: vectors, motion graphs and the equations of uniform acceleration

1. Vectors, scalars and adding vectors at any angle

A scalar has size only (distance, speed, time, mass, energy). A vector has size and direction (displacement, velocity, acceleration, force, momentum). TASC markers deduct half a mark when a vector answer has no direction, so every final vector answer needs a magnitude, a unit and a direction such as 'N 32° E', '32° east of north' or 'down the slope'.

Worked example (right angles). A boat points north at 4.0 m s⁻¹ relative to the water while the current carries it east at 2.5 m s⁻¹. Draw the vectors head to tail: the resultant is the hypotenuse. Magnitude = √(4.0² + 2.5²) = 4.72 m s⁻¹; direction = tan⁻¹(2.5/4.0) = 32.0° east of north. Writing 'tan⁻¹(4.0/2.5)' gives 58.0°, the angle measured from east, which is only correct if you say so.

Non-right-angled triangles. When the two vectors are not perpendicular, either resolve each into components and add the components, or draw the triangle and use the cosine rule a² = b² + c² − 2bc cos A for the magnitude and the sine rule for the angle. Both rules are on the Information Sheet. Components are safer when there are three or more vectors.

Change in velocity. Δv = v − u = v + (−u). A car travelling 15 m s⁻¹ north rounds a corner and travels 15 m s⁻¹ east: speed is unchanged, but Δv = √(15² + 15²) = 21.2 m s⁻¹ towards the south-east. Draw v and then −u (pointing south) head to tail. The 2024 assessment report notes that students who drew the vector triangle were far more successful than those who tried to add numbers directly.

2. Displacement–time, velocity–time and acceleration–time graphs

Three graph rules carry most of the marks: the gradient of an s–t graph is velocity, the gradient of a v–t graph is acceleration, and the area under a v–t graph is displacement (area under an a–t graph is change in velocity). A curved s–t graph means the velocity is changing; take a tangent to find the instantaneous velocity at one point and quote the gradient with units.

Worked example. A cyclist accelerates uniformly from rest to 12 m s⁻¹ in 4.0 s, rides at 12 m s⁻¹ for 8.0 s, then brakes uniformly to rest in 6.0 s. Displacement = triangle + rectangle + triangle = ½(4.0)(12) + (12)(8.0) + ½(6.0)(12) = 24 + 96 + 36 = 156 m. Average speed over the 18 s = 156/18 = 8.67 m s⁻¹, not the average of 0 and 12. Braking acceleration = (0 − 12)/6.0 = −2.0 m s⁻², the negative sign showing it is opposite to the motion.

Sketching matched graphs. For the same trip the a–t graph is three horizontal segments (+3.0, 0, −2.0 m s⁻²) and the s–t graph is a curve bending upward, then a straight line, then a curve flattening to a horizontal finish. Markers check that the s–t graph has no sharp corners where the velocity is continuous and that it never decreases while the velocity is positive.

Area below the axis. On a v–t graph, area below the time axis is displacement in the negative direction. Distance travelled adds the magnitudes of all areas; displacement adds them with signs. If a question asks 'how far from the start', it wants displacement.

3. Choosing and using the equations of uniform acceleration

The Information Sheet gives v = u + at, s = ut + ½at² and v² = u² + 2as. They apply only when acceleration is constant. Write the five symbols s, u, v, a, t, fill in the three you know, cross out the one you neither know nor want, and choose the equation without it. This one-line list earns a method mark even if the algebra later slips.

Worked example (stopping distance). A car travels at 90 km h⁻¹. The driver reacts in 0.80 s and the brakes then decelerate the car at 6.5 m s⁻². First convert: 90 ÷ 3.6 = 25 m s⁻¹. Reaction distance (constant speed) = 25 × 0.80 = 20 m. Braking distance from v² = u² + 2as with v = 0: s = 25²/(2 × 6.5) = 48.1 m. Stopping distance = 20 + 48.1 = 68.1 m. Splitting the motion into two stages is the key step; using one equation for the whole journey is a common error because the acceleration is not constant across both stages.

Doubling the speed. Because s = u²/2a for braking, doubling the initial speed quadruples the braking distance while only doubling the reaction distance. A 'justify' question about speed limits is answered with this proportional reasoning, not with a vague statement that faster cars need more room.

Units first. Convert km h⁻¹ to m s⁻¹ (÷ 3.6), km to m and minutes to seconds before substituting. The assessment reports repeatedly list unconverted units as a main source of lost marks in Section A.

Which equation when? No time given or wanted: v² = u² + 2as. No final velocity: s = ut + ½at². No displacement: v = u + at. Practising this choice on ten quick problems is more useful than rereading derivations.

4. Vertical motion under gravity and sign conventions

Near Earth's surface a freely falling object has a = g = 9.81 m s⁻² downward whether it is moving up, down or momentarily at rest. Choose a positive direction at the start (say up = positive, so a = −9.81 m s⁻²) and keep it for every quantity in the problem.

Worked example. A ball is thrown vertically up at 14 m s⁻¹ from a hand 1.5 m above the ground. Time to the top: v = u + at gives 0 = 14 − 9.81t, so t = 1.43 s. Maximum height above the hand: s = u²/2g = 14²/(2 × 9.81) = 9.99 m (11.5 m above the ground). Time to reach the ground: the ground is at s = −1.5 m, so −1.5 = 14t − 4.905t², which rearranges to 4.905t² − 14t − 1.5 = 0. The positive root is t = 2.96 s. Impact velocity v = 14 − 9.81(2.96) = −15.0 m s⁻¹, that is 15.0 m s⁻¹ downward.

Symmetry check. Without the extra 1.5 m drop, the ball would return to the hand after 2 × 1.43 = 2.86 s at 14 m s⁻¹ downward. The extra drop adds a little time and a little speed, so 2.96 s and 15.0 m s⁻¹ are sensible. Quick checks like this catch sign errors before the marker does.

Dropped objects. A stone dropped from rest falls for 2.5 s: v = 9.81 × 2.5 = 24.5 m s⁻¹ down and s = ½ × 9.81 × 2.5² = 30.7 m. At the top of a vertical throw the velocity is zero but the acceleration is still 9.81 m s⁻² down; answering 'zero acceleration at the top' is a classic wrong answer.

Up versus down. If you choose down as positive instead, a = +9.81 m s⁻², u = −14 m s⁻¹ and the ground is at +1.5 m; the numbers are the same, only the signs move. Mixing conventions within one problem is what loses marks.

5. Air resistance and terminal velocity

The equations of uniform acceleration assume no air resistance. A real falling object experiences a drag force that increases with speed. At release the only force is weight, so the acceleration is g. As speed rises, drag grows, the net force (weight − drag) falls and the acceleration decreases. When drag equals weight the net force is zero and the object falls at constant terminal velocity.

Graph shapes. A v–t graph for a skydiver starts with gradient 9.81 m s⁻², curves over and approaches a horizontal asymptote at terminal velocity. Opening the parachute suddenly increases drag beyond weight, so the net force is upward, the skydiver decelerates (the velocity still points down) and the graph curves down to a new, lower terminal velocity. An a–t graph starts at 9.81 m s⁻², falls towards zero, jumps to a large negative value when the parachute opens, then returns to zero.

Model explanation. 'Initially the only force acting is weight, so the acceleration equals g. As the speed increases, air resistance increases, reducing the net downward force and therefore the acceleration (a = F_net/m). When air resistance equals weight, the net force is zero, so the skydiver continues at constant terminal velocity (Newton's first law).' This answer names the forces, links them to the acceleration with Newton's second law and ends with the first law, which is the chain markers look for.

Mass and drag. Two balls of the same size but different mass reach different terminal velocities: the heavier ball needs a larger drag force to balance its weight, so it must fall faster before balance is reached.

6. Reading data and describing motion precisely

Questions often supply a table of position and time readings (for example from a motion sensor or video analysis) and ask whether the acceleration is uniform. Calculate successive changes in displacement over equal time intervals. If the displacement increments themselves increase by a constant amount, the acceleration is uniform: for Δt equal intervals, the difference between successive increments equals aΔt².

Example of the reasoning. A trolley's positions at 0.20 s intervals are 0.00, 0.06, 0.24, 0.54 and 0.96 m. Increments: 0.06, 0.18, 0.30, 0.42 m; the differences are all 0.12 m, so a = 0.12/(0.20)² = 3.0 m s⁻² and the motion is uniformly accelerated. Stating the constant second difference is the evidence; writing 'it looks like a curve' is not.

Precise language. Distinguish 'slowing down' (velocity and acceleration in opposite directions) from 'negative acceleration' (acceleration in the chosen negative direction). An object moving in the negative direction with a negative acceleration is speeding up. In describe questions, give the direction of velocity and acceleration separately in each stage.

Average versus instantaneous. Average velocity = total displacement ÷ total time; instantaneous velocity is the gradient of the tangent at one instant. For uniform acceleration only, the average velocity also equals (u + v)/2, which is a quick check on a v–t calculation.

Reading a gradient from a graph. Choose two widely separated points on the line or tangent, show the rise and run with units on the graph, and quote the gradient to two or three significant figures.

7. Exam technique for Section A kinematics

Structure every calculation. State the equation in symbols, substitute with units, give the answer to sensible significant figures (match the least precise data, usually 2–3) with a unit and, for vectors, a direction. Assessment reports from 2021 to 2025 deduct half marks for each missing unit or direction and for excessive significant figures, and they are consistent about this.

'Show that' items. When the answer is given (for example 'show that the ball rises 9.99 m'), you must show every substitution and carry one more significant figure than the target so the reader can see you did not work backwards. Then use the given value in later parts even if your own answer differed.

Sketch items. Label both axes with quantity and unit, mark key values (for example 12 m s⁻¹ and 4.0 s) on the axes, and make gradients and curvature consistent with the physics. A straight line where a curve belongs, or a discontinuity in velocity, loses the mark even if the general shape is right.

Time management. Section A carries 45 marks in about 45 minutes. A 3-mark kinematics part should take about three minutes; if an equation choice is not obvious within a minute, write the s, u, v, a, t list, pick an equation, and move on. Leave a gap and return rather than abandoning later questions.

Check sense. Car accelerations above about 10 m s⁻², projectile speeds faster than sound in a sport context, or a time of flight shorter than the time to reach the top are all warnings to recheck sign conventions and unit conversions.