- Calculate acceleration and tell from the signs whether a body speeds up or slows down
- Write the velocity and coordinate equations and read x₀, v₀ₓ, aₓ from an equation
- Apply the time-free formula and the 1 : 3 : 5 rule
- Solve braking, meeting and catching-up problems
The traffic light turns green. A car pulls away, and after 5 seconds the speedometer already shows 36 km/h. At the next light the car brakes and stops. Such changes of velocity are described by acceleration. In this lesson you will learn every formula of uniformly accelerated motion — velocity, coordinate, the time-free formula, the 1 : 3 : 5 rule — and use them in braking and catching-up problems.
What is acceleration?
The change in velocity divided by the time in which it happens. Acceleration is a vector and shows how much the velocity changes every second. Its unit is m/s². If the acceleration stays constant, the motion is uniformly accelerated. Acceleration is measured by an accelerometer — every smartphone has one.
- aₓacceleration projection, in m/s²
- v₀ₓ, vₓprojections of the initial and final velocity, in m/s
- ttime of the change, in s
If vₓ and aₓ have the same sign, the speed grows; if the signs are opposite, the body slows down. So a “negative acceleration” does not always mean braking!
a) A car starts from rest and reaches 20 m/s in 8 s. b) A bus moving at 15 m/s brakes and stops after 5 s. c) A body has v₀ₓ = −10 m/s and aₓ = +2 m/s². Is it speeding up or slowing down? What is vₓ after 8 s?
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b) a = (0 − 15) / 5 = −3 m/s² — the speed drops by 3 m/s every second.
c) The signs are opposite → the body is slowing down; vₓ = −10 + 2t, it stops at t = 5 s and then speeds up along the axis: vₓ(8) = −10 + 16 = 6 m/s.
The velocity and coordinate equations
- vₓvelocity projection at time t
Velocity depends linearly on time.
- x₀initial coordinate, in m
- xcoordinate at time t, in m
The coordinate equation of uniformly accelerated motion. The displacement is sₓ = x − x₀ = v₀ₓt + aₓt²/2; if the body does not turn back, this is also the distance.
- (v₀ₓ + vₓ) / 2(v₀ₓ + vₓ) / 2average velocity in uniformly accelerated motion
The average velocity is the mean of the initial and final velocities — only for uniformly accelerated motion!
The meaning of these formulas is visible on a v–t graph: the velocity grows linearly, so the shape under the graph is a trapezium whose area is the average velocity times the time. Putting vₓ = v₀ₓ + aₓt gives (v₀ₓ + v₀ₓ + aₓt)/2 · t = v₀ₓt + aₓt²/2 — exactly the displacement in the coordinate equation.
A body’s coordinate changes as x = 5 − 4t + 2t² (SI). Find x₀, v₀ₓ and aₓ and write the velocity equation. When and where does the body stop? What are the displacement and the distance in 3 s?
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vₓ = −4 + 4t; vₓ = 0 → t = 1 s, x(1) = 5 − 4 + 2 = 3 m.
x(3) = 5 − 12 + 18 = 11 m → displacement 11 − 5 = 6 m.
Distance: first from 5 m to 3 m (2 m), then from 3 m to 11 m (8 m) → 10 m.
Leyla starts cycling from rest with an acceleration of 0.5 m/s². What is her velocity after 10 s, how far does she travel in that time and what is her average speed?
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s = a · t² / 2 = 0.5 · 100 / 2 = 25 m.
Average speed: 25 / 10 = 2.5 m/s = (0 + 5) / 2 — the average-velocity formula checks out.
The time-free formula
When a problem neither gives nor asks for the time, substitute t = (v − v₀) / a into the average-velocity formula: s = (v₀ + v) / 2 · (v − v₀) / a. This gives a formula without time.
- v₀, vinitial and final velocity, in m/s
- aacceleration (negative when braking), in m/s²
Another form: v² − v₀² = 2as. Braking distance: s = v₀² / (2|a|).
1) A plane needs 60 m/s to take off. Starting from rest with an acceleration of 2.5 m/s², what runway length does it need?
2) A car moving at 20 m/s brakes with a deceleration of 5 m/s². What is its braking distance?
3) A body moving at 10 m/s speeds up at 3 m/s². What is its velocity after 50 m?
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2) s = (0 − 20²) / (2 · (−5)) = 40 m.
3) v² = 10² + 2 · 3 · 50 = 400 → v = 20 m/s.
| Which quantity is missing from the problem? | Formula |
|---|---|
| time t | s = (v² − v₀²) / (2a) |
| acceleration a | s = (v₀ + v) / 2 · t |
| final velocity v | s = v₀ · t + a · t² / 2 |
| distance s | v = v₀ + a · t |
The 1 : 3 : 5 rule and the distance in the n-th second
In uniformly accelerated motion from rest, the distances covered in t, 2t, 3t, … grow as 1 : 4 : 9 : … (as the square of time). The distances covered in successive equal intervals are in the ratio of odd numbers: 1 : 3 : 5 : 7 : … Galileo discovered this.
- sₙdistance covered in the n-th second by a body starting from rest, in m
- nnumber of the second
It comes from the difference sₙ = s(n) − s(n − 1).
1) A body starts from rest with an acceleration of 4 m/s². How far does it go in the 3rd second?
2) A body accelerating uniformly from rest covers 2 m in the first second. How far does it go in the 5th second, and in the first 5 seconds?
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2) Second by second the ratio is 1 : 3 : 5 : 7 : 9 → in the 5th second 2 · 9 = 18 m. In the first 5 s the ratio is 1 : 25, so 2 · 25 = 50 m.
A body accelerating uniformly from rest covers 14 m in the 4th second. Calculate its acceleration.
A) 4 m/s² B) 3.5 m/s² C) 2 m/s² D) 7 m/s² E) 1.75 m/s²
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E (1.75) wrongly takes 14 m as the whole distance of the first 4 seconds, and B (3.5) just divides 14 by 4.
Braking and catching-up problems
A braking body does not go backwards after it stops. So first find the stopping time (ts = v₀ / |a|): if the time asked for is longer, the distance is just the braking distance. For meeting and catching-up problems, write a coordinate equation for each body on the same axis and solve x₁ = x₂.
A car moving at 20 m/s brakes with a deceleration of 4 m/s². Find the distance it travels in 6 s.
A) 48 m B) 50 m C) 72 m D) 120 m E) 40 m
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s = v₀² / (2a) = 400 / 8 = 50 m (B).
A (48 m) comes from putting t = 6 s into s = v₀t − at²/2, which pretends the car rolls backwards.
A car moving at 20 m/s brakes at 4 m/s² until it stops. Find the distance it covers in each second. How far does it go in the last second before stopping?
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So second by second: 18, 14, 10, 6, 2 m; in the last second the distance is a/2 = 2 m.
Total 18 + 14 + 10 + 6 + 2 = 50 m — the same as the braking distance.
A motorcycle starts from rest at point A with an acceleration of 4 m/s². At the same moment a cyclist leaves point B, 100 m away, riding towards it at a constant 10 m/s. When and how far from A do they meet?
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x₁ = x₂ → 2t² + 10t − 100 = 0 → t² + 5t − 50 = 0 → (t + 10)(t − 5) = 0 → t = 5 s.
x = 2 · 25 = 50 m. The motorcycle’s speed is then 4 · 5 = 20 m/s.
As a bus pulls away from a stop with an acceleration of 1 m/s², a cyclist passes it in the same direction at a constant 5 m/s. After how many seconds does the bus catch the cyclist?
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Answer: 10. Check: the bus is then at 10 m/s = 2 · 5 m/s, and they meet 50 m from the stop.
- 1.36 km/h = m/s
- 2.v₀ = 0, a = 3 m/s², t = 4 s → v = m/s
- 3.From rest: s₁ : s₂ : s₃ = 1 : 3 :
- 4.v₀ = 10 m/s, a = −5 m/s² → braking distance m
Key points
- aₓ = (vₓ − v₀ₓ) / t; if vₓ and aₓ have the same sign the body speeds up, if opposite it slows down.
- vₓ = v₀ₓ + aₓt, x = x₀ + v₀ₓt + aₓt²/2, average velocity (v₀ + v)/2.
- Time-free formula: s = (v² − v₀²) / (2a); braking distance is proportional to v₀².
- From rest: distances in successive seconds go 1 : 3 : 5 : …, sₙ = a(2n − 1)/2.
- When braking, find the stopping time first; for a meeting, solve x₁ = x₂.
Check yourself
12 questions. Every correct answer earns XP.