What is the acceleration?
Fill any three of starting speed, final speed, time and acceleration. The acceleration calculator works out the fourth and the distance covered.
- Acceleration
- 3.3528
Going from 0 mph to 60 mph in 8 s is an average acceleration of 3.3528 m/s².
- Acceleration in gg
- 0.3419
- Acceleration in ft/s²ft/s²
- 11
- Distance covered
- 352 ft
Acceleration: 3.3528. Going from 0 mph to 60 mph in 8 s is an average acceleration of 3.3528 m/s².
How to calculate
Computes average acceleration from the starting speed, the final speed and the time, or any one of the four from the other three, with the distance covered and the acceleration in g.
Example with the default inputs (Starting speed 0 mph, Final speed 60 mph, Time 8 s): Going from 0 mph to 60 mph in 8 s is an average acceleration of 3.3528 m/s².
Formula: a = (v − u) ÷ t; v = u + a × t; u = v − a × t; t = (v − u) ÷ a; distance = (u + v) ÷ 2 × t.
- Motion is along a straight line, and the speeds are in the direction of travel (0 or more).
- The acceleration is the average over the time; the distance assumes it is constant.
- Units are converted exactly: 1 mph = 0.44704 m/s, 1 ft = 0.3048 m, standard gravity g = 9.80665 m/s² (NIST SP 811).
Worked examples
Each example is checked against the calculator on every build.
- Starting speed 0 mph, Final speed 60 mph, Time 8 s gives Acceleration (m/s²) 3.3528, Acceleration in g 0.34189, Distance covered 352 ft.Source: OpenStax University Physics Volume 1, section 3.4, Motion with Constant Acceleration (https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration, retrieved 2026-10-01); NIST SP 811: 1 mph = 0.44704 m/s
- Starting speed 156.6 mph, Final speed 0 mph, Acceleration (m/s²) -5 gives Time 14 s, Distance covered 1,607.61 ft.Source: OpenStax University Physics Volume 1, section 3.4, Motion with Constant Acceleration, equation v = v₀ + at (https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration)
- Starting speed 22.37 mph, Acceleration (m/s²) 2, Time 5 s gives Final speed 44.74 mph, Distance covered 246.063 ft, Acceleration in ft/s² 6.56168.Source: OpenStax University Physics Volume 1, section 3.4, Motion with Constant Acceleration (https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration)
How it works
Fill any three of the four boxes and leave one empty. With u the starting speed, v the final speed, t the time and a the acceleration:
- a = (v − u) ÷ t
- v = u + a × t
- u = v − a × t
- t = (v − u) ÷ a
The calculator converts every value to SI units first (speeds in m/s, time in seconds, acceleration in m/s²), solves, and then shows:
- Acceleration in m/s², to 6 significant figures.
- Acceleration in g: a ÷ 9.80665, to 4 significant figures.
- Acceleration in ft/s²: a ÷ 0.3048, to 6 significant figures.
- Distance covered: (u + v) ÷ 2 × t, to 6 significant figures, in feet for US units and metres for Metric. This is the distance at constant acceleration. It is computed exactly from the speeds and time as you typed them (each read as the decimal in its unit), so a value that ends in 5 at the 7th figure rounds up (1,793.785 m shows as 1,793.79 m). In US units the page converts those exact metres to feet in double precision for display, so a value that ends in 5 at the 7th figure in feet can round down (2,286.195 ft can show as 2,286.19 ft).
Values are rounded half up for display only.
Rules
- Speeds are 0 or more (up to 10,000,000 m/s), measured along the direction of travel; the object does not turn back.
- The time is more than 0 and at most 10,000 seconds.
- The acceleration is from −10,000 to 10,000 m/s²; it is negative when the object slows down.
- A value that would come out of these ranges gives no answer. For example, solving for the time with an acceleration of 0, or with a speed change of the opposite sign to the acceleration, has no answer.
- Unit conversions are exact: 1 mph = 0.44704 m/s, 1 km/h = 1 ÷ 3.6 m/s, 1 ft/s = 0.3048 m/s, 1 knot = 1,852 ÷ 3,600 m/s.
Worked examples by hand
0 to 60 mph in 8 seconds. 60 mph = 60 × 0.44704 = 26.8224 m/s. a = (26.8224 − 0) ÷ 8 = 3.3528 m/s², which is 3.3528 ÷ 9.80665 = 0.3419 g. Distance = (0 + 26.8224) ÷ 2 × 8 = 107.2896 m.
Braking from 70 m/s to a stop at −5 m/s². t = (0 − 70) ÷ (−5) = 14 s. Distance = (70 + 0) ÷ 2 × 14 = 490 m.
10 m/s, accelerating at 2 m/s² for 5 seconds. v = 10 + 2 × 5 = 20 m/s. Distance = (10 + 20) ÷ 2 × 5 = 75 m. In ft/s², 2 ÷ 0.3048 = 6.56168.
Other questions people ask
How do I calculate acceleration?
Subtract the starting speed from the final speed and divide by the time: a = (v − u) ÷ t. A car going from 0 to 60 mph (26.8224 m/s) in 8 seconds accelerates at 26.8224 ÷ 8 = 3.3528 m/s².
What are the units of acceleration?
Speed per unit of time. In SI units it is metres per second squared (m/s²): the speed changes by that many metres per second every second. The calculator also shows feet per second squared and g.
What does negative acceleration mean?
The object slows down: the final speed is lower than the starting speed. Braking from 70 m/s to a stop in 14 seconds is an acceleration of −5 m/s², often called a deceleration of 5 m/s².
What is acceleration in g?
The acceleration divided by standard gravity, 9.80665 m/s². 1 g is the acceleration of a dropped object near the Earth’s surface, ignoring air. 0 to 60 mph in 8 seconds is about 0.34 g.
How do I find the time from the acceleration?
Rearrange the formula: t = (v − u) ÷ a. Fill the two speeds and the acceleration, and leave the time empty.
How far does the object travel while it accelerates?
At constant acceleration, the distance is the average of the two speeds times the time: d = (u + v) ÷ 2 × t. From 0 to 26.8224 m/s in 8 seconds that is 13.4112 × 8 = 107.29 m.
Is this the same as instantaneous acceleration?
Only when the acceleration is constant. Otherwise the result is the average acceleration over the time, which says nothing about the peak.