Acceleration Calculator
Acceleration from a change in velocity over a time, or any two of initial velocity, final velocity, acceleration, displacement and time from the other three.
Result
Acceleration: 11 ft/s²
Constant acceleration in a straight line. Real starts and stops rarely hold one acceleration throughout.
- Final velocity
- 60 mph (88 ft/s)
- Initial velocity
- 0 mph (0 ft/s)
- Displacement
- 352 ft
- Time
- 8 s
- Average velocity
- 30 mph (44 ft/s)
https://calcgallery.com/physics/acceleration-calculator/
Results come from the published equations applied to the values you enter. They are not electrical design, conductor sizing, laboratory safety or medical guidance. Weather indices describe a standard person in standard conditions and are not a forecast. Where codes apply, local rules vary and must be checked. Read the full disclaimer
How to use the Acceleration Calculator
- Choose the three values you know, for example initial velocity, final velocity and time.
- Enter them with a sign that shows direction. Take one direction as positive, and make a velocity or acceleration the other way negative.
- Read the acceleration, or whichever value you were missing, with the displacement, time and average velocity that go with it.
The formulas
v = u + ats = ut + ½at²v² = u² + 2ass = (u + v) ÷ 2 × ta = (v − u) ÷ tu is the initial velocity, v the final velocity, a the acceleration, s the displacement and t the time. The page rearranges whichever pair of these it needs for the three values you know, and when a time comes from a quadratic it takes the first positive root.
Limits
The acceleration is taken as constant for the whole time, and the motion as a straight line. Displacement is the change in position, not the distance travelled: if the velocity changes sign on the way, the object has turned round and the two differ, which the note points out. Air resistance and friction are not modelled; they are part of whatever acceleration you enter.
With your numbers
- a = (v − u) ÷ t = (88 ft/s − 0 ft/s) ÷ 8 s = 11 ft/s²
- s = (u + v) ÷ 2 × t = 352 ft
Quick reference
Constant acceleration in a straight line. Real starts and stops rarely hold one acceleration throughout.
| Example | Result |
|---|---|
| 0 to 60 mph in 8 s | Acceleration: 11 ft/s² |
| 0 to 100 km/h in 8 s | Acceleration: 3.472 m/s² |
| Landing, 70 m/s at −1.5 m/s² for 40 s | Final velocity: 10 m/s |
| Dragster, 26 m/s² from rest for 5.56 s | Final velocity: 144.6 m/s |
| Stopping from 30 m/s at −6 m/s² | Displacement: 75 m |
| Stopping from 60 mph in 150 ft | Acceleration: -25.81 ft/s² |
| Time to cover 402 m at 26 m/s² from rest | Final velocity: 144.6 m/s |
| 100 m in 5 s at 2 m/s² | Initial velocity: 15 m/s |
Common questions about the Acceleration Calculator
- How do I work out acceleration from speed and time?
- Divide the change in velocity by the time it took: a = (v − u) ÷ t. Going from rest to 60 mph, which is 88 ft/s, in 8 seconds is 88 ÷ 8 = 11 ft/s², about 3.35 m/s². The distance covered at that steady acceleration is the average velocity times the time, 44 ft/s × 8 s = 352 ft.
- What are the SUVAT equations?
- Four equations for motion in a straight line at constant acceleration, named after the letters for displacement s, initial velocity u, final velocity v, acceleration a and time t. Each one leaves out one of the five: v = u + at has no s, s = ut + ½at² has no v, v² = u² + 2as has no t, and s = (u + v)t ÷ 2 has no a. Knowing any three lets you find the other two.
- Why is my acceleration negative?
- The sign gives the direction. If the object is moving the positive way and slowing down, its acceleration points backwards and is negative. An airliner landing at 70 m/s and braking at −1.5 m/s² for 40 s is down to 10 m/s, having rolled 1,600 m, the figures in OpenStax University Physics example 3.7.
- How long does it take to stop, and how far does it go?
- Choose initial velocity, final velocity and acceleration, with a final velocity of zero and a negative acceleration. From 30 m/s at −6 m/s² the page gives 5 s and 75 m. The acceleration has to point against the motion, or the object never stops, and the page says so.
- Can the displacement be reached twice?
- Yes. A ball thrown upwards passes each height once going up and once coming down, and something that turns around can come back past its start. When you give the displacement and the acceleration, the page takes the first time it gets there.