Article
A car pulls away from traffic lights, a bike speeds up, a dropped ball falls faster and faster. Acceleration stays constant, so speed grows at a steady rate. Below are the formulas for distance and final speed.
Part 2 of Physics without mysteries. If you would rather start with constant speed, read uniform motion. The next topic is braking.
On this page: v₀ is initial speed, vₖ is final speed, a is acceleration, s is distance, t is time. ½at² means one-half times a times t squared.
When a car pulls away from traffic lights, it moves slowly at first and then faster and faster. It does not jump straight to 90 km/h. How quickly its speed increases is called acceleration.
Acceleration tells you how much your speed increases each second. The unit is meters per second per second, written m/s². It looks strange at first, but the idea is simple. If a = 2 m/s², then after 1 s you are 2 m/s faster than at the start, after 2 s you are 4 m/s faster, and after 3 s you are 6 m/s faster, provided the acceleration stays the same.Acceleration tells you how much your speed increases each second. The unit is feet per second per second, written ft/s². It looks strange at first, but the idea is simple. If a = 2 m/s², then after 1 s you are 2 m/s faster than at the start, after 2 s you are 4 m/s faster, and after 3 s you are 6 m/s faster, provided the acceleration stays the same.
In practice, we usually use a simple special case: uniformly accelerated motion. Here “uniform” does not mean constant speed. It means constant acceleration. The speed can still rise; it just rises at a steady rate.
This model works well for many short situations: pulling away from lights, speeding up on a bike, or the first few seconds of a short free fall. The formulas are short. You can work them out on paper or with the accelerated-motion calculator.
Before the formulas, you need three quantities: initial speed v0 (how fast you are already moving at the start), acceleration a (how quickly that speed increases), and time t (how long this lasts). From those three you can find the distance s and the final speed vk.
The first formula is straightforward. With constant a, the final speed equals the initial speed plus the gain each second multiplied by the number of seconds:
vk = v0 + a·t
If you start from rest, v0 = 0 and the formula becomes vk = a·t. With a = 3.5 m/s² and t = 8 s you get vk = 28 m/s, or about 100 km/h. That is a typical figure for a car accelerating from a standstill.
The second formula gives the distance. It contains ½at². Why the half? With constant acceleration, speed rises evenly from v0 to vk. So for distance you use the average speed (v0 + vk)/2, not the final speed alone. Distance is always average speed times time.
Put vk = v0 + a·t into that average and you get:
s = v0·t + ½·a·t²
When v0 = 0, this simplifies to s = ½·a·t². With a = 3.5 m/s² and t = 8 s, t² = 64, so s = 0.5 · 3.5 · 64 = 112 m. That is roughly the length of a soccer field. The same numbers appear in the example below: work it out by hand, then check the answer in the calculator.
A common mistake looks like this: “I reached 28 m/s and I was moving for 8 s, so the distance is 28 · 8.” That answer is too large. For most of the time you were slower than at the end, so you must use the average speed, not the final speed. On a speed-time graph the plot is a straight line sloping up, and the distance is the area under that line (a triangle or a trapezoid), not a rectangle of height vk.
Acceleration comes from forces. Newton’s second law is F = m·a: the net force equals mass times acceleration. Rearranged, a = F/m, so for the same force a larger mass gives a smaller acceleration. The engine pushes the car forward; friction and air resistance push back. Over a short stretch we often treat the net force as roughly constant, so a is constant too.
Keep two questions separate. If you only know how much the speed changed and how long it took, first find a with the acceleration calculator (a = Δv/t). If you already have a, together with v0 and t, use uniformly accelerated motion to find the distance and vk. If you know the mass and the force, use Newton’s second law F = m·a and find a = F/m.
Units matter. Use speed in m/s, time in seconds, acceleration in m/s², and distance in meters. If your speed is in km/h, divide by 3.6. For example, going from 0 to 100 km/h is ≈ 27.8 m/s. Over 8 s that gives a ≈ 3.5 m/s². Only then put that value of a into the distance formula. If you type 100 as though it were already in m/s, the result will be nonsense.Units matter. Use speed in ft/s, time in seconds, acceleration in ft/s², and distance in feet. If your speed is in mph, convert to ft/s (multiply by about 1.467). For example, going from 0 to 100 km/h is ≈ 27.8 m/s. Over 8 s that gives a ≈ 3.5 m/s². Only then put that value of a into the distance formula. If you type an mph value as though it were already in ft/s, the result will be nonsense.
Braking uses the same formulas; only the speed decreases. You can treat this as negative acceleration or, more clearly at first, as positive deceleration. When a car has to stop, a stopping-distance formula is useful. We cover that in the separate article on braking, so you do not mix up the signs before the speeding-up case is clear.
There is also a formula that does not need time: vk² = v0² + 2·a·s. Use it when you know the distance and the speeds, but not the time. This is not a different topic. It is the same constant a, rearranged for different known values.
This model breaks down when the acceleration is not constant: in a strong wind, through mud, when air resistance grows with speed, or in a turn where the direction of motion changes. Then ½at² is only a rough estimate, or the wrong tool. For everyday estimates of pulling away and short bursts of acceleration, constant a is good enough.
Before you start, check these points in order: (1) whether you can treat a as constant, (2) whether the units match the choice in the page header, (3) whether v0 is zero, (4) whether you need the formula for vk or for s, (5) whether the answer looks sensible (a bicycle is not doing 300 m/s). Then check the arithmetic in the calculator.
If you have already read the article on uniform motion, compare the graphs. With constant speed, distance grows in a straight line. With constant acceleration, distance grows faster and faster: later seconds cover more ground than earlier ones. That is why time is squared in ½at².
Open uniformly accelerated motion. Enter the initial speed v0 (use 0 if you start from rest), the acceleration a, and the time t. Read the distance s and the final speed vk.
If you only know that the speed went from v₁ to v₂ in time t, first calculate a = (v₂ − v₁)/t in the acceleration calculator, then come back here with v0 = v₁. Do not find the distance by multiplying v₂ by t.
We treat acceleration as constant over the stretch we are calculating. Motion is in a straight line, or we look at one direction only. We work in meters, seconds and m/s. You do not need force or mass unless you want to know where a comes from.We treat acceleration as constant over the stretch we are calculating. Motion is in a straight line, or we look at one direction only. We work in feet, seconds and ft/s. You do not need force or mass unless you want to know where a comes from.
Use something else when a clearly changes, when turning and change of direction matter, when air resistance depends strongly on speed, or when you include the driver’s reaction time in a full stop. In those cases the braking article and calculators are a better fit than forcing ½at².
| You know | You want | Formula |
|---|---|---|
| v₀, a, t | s | s = v₀t + ½at² |
| v₀, a, t | vₖ | vₖ = v₀ + at |
| Δv, t | a | a = Δv/t (separate calculator) |
| F, m | a | F = m·a → a = F/m → distance |
A car starts at v₀ = 0 with acceleration a = 3.5 m/s² for t = 8 s. What distance does it cover and what is vₖ?
Steps
Answer: s = 112 m, vₖ = 28 m/s
Speed tells you how fast you are moving (for example 20 m/s). Acceleration tells you how quickly that speed is changing (for example by 2 m/s every second). You can move fast with zero acceleration: just hold a steady speed on the freeway.
Because for most of the time you were slower than at the end. We use the average of the initial and final speeds. After substituting vₖ = v₀ + at, that average produces the ½at² term next to v₀t.
Only when you start from rest, so v₀ = 0. If you already had some speed and then accelerated, you must also add v₀·t. Leaving out v₀ is a common mistake.
Convert 100 km/h to m/s: ≈ 27.8 m/s. Then a ≈ 3.5 m/s². With that a you can find the distance covered in those 8 seconds when v₀ = 0.Convert 100 km/h to ft/s: ≈ 27.8 m/s. Then a ≈ 3.5 m/s². With that a you can find the distance covered in those 8 seconds when v₀ = 0.
Yes. Negative a means the speed is falling in your chosen direction, so you are braking or slowing down. At first, it is often clearer to talk about positive deceleration and read the braking article.
From Newton’s second law: F = m·a, so a = F/m. Only then do a, time and v₀ give distance. Do not multiply force by time to get a distance in meters or feet.
If you are asking “what is a?”, use the acceleration calculator. If you already know a (plus v₀ and t) and want distance or vₖ, use accelerated motion. They are two steps in the same calculation.
With constant a it is a straight line. The slope of that line is a. The area under the line from the start to time t is the distance. A rectangle of height vₖ gives too large an answer, because it pretends you had that speed the whole time.
Yes, if you can treat the acceleration as roughly constant over a short stretch. The same formulas apply to a lift at about 2 m/s²; only the situation is different.
Compare it with everyday experience. a ≈ 3 m/s²-4 m/s² for a few seconds gives car-like speeds, not orbital ones. If a calculation for a bike gives hundreds of m/s, you have almost certainly mixed up the units (typed km/h as if they were m/s).Compare it with everyday experience. a ≈ 3 m/s²-4 m/s² for a few seconds gives car-like speeds, not orbital ones. If a calculation for a bike gives hundreds of ft/s, you have almost certainly mixed up the units (typed mph as if they were ft/s).