Equations of motion: how v = u + at, s = ut + ½at² and v² = u² + 2as fit together

Guide · Updated 2026-11-10

The three equations of motion for steady acceleration, what each symbol means, and how to see them on a velocity-time graph. Solve a problem from three known values, and find a car's stopping distance.

Open the Equations of motion Free, in your browser, nothing uploaded.

Three equations describe any object that speeds up or slows down at a steady rate in a straight line. They link the starting speed, the final speed, the acceleration, the time and the distance. The equations of motion tool shows them on a velocity-time graph, so you can see where each one comes from. Before this, the motion graphs live tool covers reading the graphs themselves. After it, projectile drag takes the same equations into two directions, and video motion tracker lets you measure a real motion from a video.

What it is

Use these symbols:

• u is the starting velocity, in metres per second (m/s).
• v is the final velocity (m/s).
• a is the acceleration (m/s²). A negative value means slowing down, if the direction you call positive is the direction of motion.
• t is the time (s).
• s is the displacement (m), the change in position in a straight line.

On a velocity-time graph, the line starts at u and rises or falls to v at time t. The area under the line is the displacement. Split that area into a rectangle and a triangle:

• The rectangle has width t and height u, so its area is u × t. This is the distance you would cover at the starting speed.
• The triangle has width t and height v − u, which is at, so its area is ½ × t × at = ½at². This is the extra distance from speeding up.

Add them up and you get s = ut + ½at². Because the rise v − u equals at, the slope of the line gives you the first equation, v = u + at. Eliminating the time t from the two gives the third, v² = u² + 2as. The third one is handy when no time is given.

Stopping distance is the same picture. The flat part is the thinking distance: speed × reaction time, before the brakes work. The slope is the braking distance, given by v² ÷ (2 × deceleration).

How to use it

1. Choose a level. Class 9 explains where the equations come from with the graph. Class 11 derives them with calculus.
2. In Explore the graph, drag the handles to change the starting velocity and the end of the line. The sliders make finer changes. Watch the three equations update with your numbers.
3. Press Play to see the car move while a dashed line travels along the graph.
4. In Solve a problem, enter any three of u, v, a, s and t, and leave the other two blank. The page shows which equations it used.
5. In Stopping distance, set the speed, the reaction time and the road surface, and read the thinking and braking distances.

A worked example

A cyclist starts at u = 5 m/s and accelerates at a = 2 m/s² for t = 6 s.

• Final speed: v = u + at = 5 + 2 × 6 = 17 m/s
• Displacement: s = ut + ½at² = 5 × 6 + ½ × 2 × 36 = 30 + 36 = 66 m
• Check with the third equation: v² = 289, and u² + 2as = 25 + 2 × 2 × 66 = 289. The answers agree.

Now a car with u = 12 m/s slows down to v = 0 with a = −3 m/s².

• Time: t = (v − u) ÷ a = (0 − 12) ÷ (−3) = 4 s
• Distance: s = (v² − u²) ÷ (2a) = (0 − 144) ÷ (−6) = 24 m

A ball is thrown upwards at 15 m/s. Taking upwards as positive, a = −10 m/s², and it has risen s = 10 m. Solving 10 = 15t − 5t² gives t = 1 s and t = 2 s. At 1 s the ball is still going up at 5 m/s. At 2 s it is coming down at −5 m/s, passing the same height on the way down.

Finally, a car at 36 km/h (10 m/s) on a wet road, with a reaction time of 0.8 s and deceleration of 4.5 m/s²: the thinking distance is 10 × 0.8 = 8 m, the braking distance is 10² ÷ 9 = 11.1 m, and the stopping distance is 19.1 m. At 72 km/h the thinking distance doubles to 16 m, but the braking distance becomes four times as long at 44.4 m, so the total is 60.4 m.

Where it is used

• Road safety. Driving schools and road authorities use stopping distances to teach why speed matters.
• Sport. Coaches estimate how fast a sprinter accelerates from the first few metres.
• Physics labs. Ticker-tape timers and light gates record the motion, and these equations check the results.
• Lifts and trains. Engineers use them to plan how fast a lift can speed up and stop comfortably.

Common mistakes

1. Mixing units. Convert km/h to m/s by dividing by 3.6 before using the equations. 54 km/h is 15 m/s, not 54.
2. Losing track of the sign. Choose a positive direction and stick to it. If forwards is positive, slowing down gives a negative acceleration.
3. Reading s as distance. s is displacement. If the object turns back, the distance travelled is larger than s.
4. Using the equations for changing acceleration. They work only for steady acceleration. For a changing one, use the area under the graph.
5. Forgetting the reaction time. A stopping distance without the thinking distance is too short, because the driver needs time to react before the brakes start.

Questions people ask

Which equation do I use if I do not know the time? Use v² = u² + 2as. It links speed and distance without t.

Do these work for falling objects? Yes, with a = g, about 9.8 m/s² downward, if you take downward as positive. The ball example above shows the same idea with the sign reversed.

Why does doubling my speed more than double the stopping distance? The thinking distance is speed × reaction time, so it doubles. The braking distance uses v², so it goes up four times. The total grows by a factor between two and four.