Applied Maths: Uniform Acceleration (Strand 2) — Higher Level Notes
The four equations of motion, time-velocity graphs, and the two-object problems that become simultaneous equations — plus the average-speed trap that catches nearly everyone once.
1. The four equations, and when each one saves you
Uniform acceleration means the acceleration never changes. Four equations follow, and picking the right one is most of the skill:
v = u + at | s = ut + ½at2 | v2 = u2 + 2as | s = ((u + v)/2)t
Read the question for what is missing. No time given? Use v2 = u2 + 2as. No final velocity? Use s = ut + ½at2. Choosing on that basis rather than by habit is worth minutes across a paper.
2. Time-velocity graphs
On a velocity-time graph the slope is the acceleration and the area under the line is the displacement. Whole questions are often just a trapezium with its area set equal to a given distance.
Watch out — area below the axis
When velocity goes negative the area still counts, but it counts negatively for displacement while counting positively for distance. A particle at +6 m/s for 5 s then -4 m/s for 10 s travels 70 m but finishes 10 m behind its start.
3. Two lines crossing does not mean what you think
With two cars on the same velocity-time axes, the crossing point is where their speeds are equal. It is not where they have gone the same distance — that needs equal areas, and it always happens later.
4. Problems that become simultaneous equations
Given two pieces of information about one motion, write s = ut + ½at2 twice and solve for u and a together.
The reading matters. "It travels 84 m in the first 6 s and 200 m in the first 10 s" means both distances run from the same point, so the second equation takes t = 10 and s = 200 — not the four seconds in between.
The average-speed trap
A particle covers 30 m in 3 s. Its speed at the start is not 30/3 = 10 m/s.
That is the average speed over the interval, and under uniform acceleration the average equals the instantaneous speed at the midpoint of the time interval — not at its start.
5. Motion under gravity
Vertical motion is the same four equations with a = -g. At Higher Level take g = 9.8 m/s2 unless a question says otherwise; Ordinary Level uses 10.
Thrown straight up with speed u, the greatest height is:
h = u2 / 2g
The 2 comes from v2 = u2 - 2gh with v = 0 at the top. Dropping it doubles your answer — the most frequent error in this section.
Two more facts worth knowing cold: time up equals time down, and the speed on return equals the speed of projection. Speed grows as the square root of the height fallen, so half the drop gives 1/√2 of the final speed, not half.
6. Two objects, one road
Write a displacement expression for each, then set them equal. A car passing at a constant 20 m/s and a motorbike starting from rest at 4 m/s2 meet when 20t = ½(4)t2, so t = 0 or t = 10. The t = 0 root is the moment they set off together — real, but not the answer.
7. Letters instead of numbers
The hardest questions here have no numbers at all. The method does not change, only your confidence. Subtracting the distance equations for two consecutive equal intervals cancels the ½ completely and leaves s2 - s1 = at2. Carrying that ½ into the answer is the classic slip.
Practise this chapter
QuizPerCard has Uniform Acceleration as concept cards, a formula reference and multiple-choice practice — every question with the full derivation, and an explanation of why each wrong option is wrong. Open the practice.