When an object starts with some velocity and then speeds up or slows down at a steady rate, its position no longer grows in a straight line with time. It follows a curve. The equation s = ut + ½at² gives the displacement at any moment, and this calculator solves it for any one of its four quantities. It also draws the position–time curve, so you can see how the motion unfolds instead of getting a single number.
How to use the s = ut + ½at² calculator
- Under Solve for, choose Displacement, Initial velocity, Acceleration or Time.
- Fill in the three remaining fields: Displacement (s), Initial velocity (u), Acceleration (a) and Time (t). Acceleration can be entered in m/s², ft/s², g, km/h per second or mph per second.
- Set Show the result in to the unit you want the answer in.
- The result also lists the final velocity (v = u + at), the average velocity (s ÷ t) and a chart of displacement over time, drawn in the length unit you are using.
The formula and its rearrangements
a = 2(s − ut) ÷ t2
t = [−u ± √(u2 + 2as)] ÷ a
Choose a positive direction. Displacement, initial velocity and acceleration are signed along that axis. Time is positive. When a = 0 the time formula reduces to t = s ÷ u.
Where the formula comes from
On a velocity–time graph, constant acceleration is a straight line rising from u to u + at. The area under that line is the displacement. It splits into a rectangle of height u and width t, worth ut, plus a triangle of base t and height at, worth ½at². The two terms of the formula are those two areas.
Worked example
Given: u = 5 m/s, a = 2 m/s², t = 4 s.
s = 5 × 4 + ½ × 2 × 4² = 20 + 16 = 36 m
Final velocity: v = 5 + 2 × 4 = 13 m/s. Average velocity: 36 ÷ 4 = 9 m/s.
The chart shows a parabola that gets steeper as time goes on. The slope at any point is the velocity, which climbs from 5 m/s to 13 m/s. Notice that the 9 m/s average is exactly halfway between those two values, which only happens because the acceleration is constant.
Solving for time: one root or two
Because t appears squared, finding the time means solving a quadratic. Suppose a cart is pushed up a long ramp at u = 10 m/s and slows at a = −4 m/s². When is it 10 m up the ramp?
t = [−10 ± √(10² + 2 × (−4) × 10)] ÷ (−4) = [−10 ± √20] ÷ (−4)
Roots: t = 1.382 s and t = 3.618 s
Both roots are positive. The cart reaches the 10 m mark at 1.382 s moving up the ramp at 4.47 m/s. It stops at its highest point, 12.5 m, at t = 2.5 s. It then rolls back and passes the 10 m mark again at 3.618 s, moving down at 4.47 m/s. The calculator reports the first time and mentions the second in a note. A ball thrown upward behaves the same way, with a = −9.80665 m/s² (standard gravity, g₀, fixed by the CGPM in 1901).
Some requests have no answer, and the calculator says so instead of returning nonsense:
- Past the turning point. Asking when the cart reaches 15 m gives u² + 2as = −20, so the square root is not real. The cart turns back at 12.5 m and never gets there.
- Only in the past. If both roots are negative, the object was at that position only before t = 0.
- No motion. With u = 0 and a = 0 the object never moves, so no time works.
Reading the position–time chart
The chart plots 13 evenly spaced points from t = 0 to the time you entered. A positive acceleration makes the curve bend upward. A negative one makes it bend downward. If the velocity changes sign during the interval, the curve rises to a peak and turns back. Try u = 10 m/s, a = −4 m/s² and t = 5 s. The displacement is 0 m, because the cart has returned to its starting point, but it actually traveled 25 m. The calculator flags this case and gives the moment the velocity changes sign (2.5 s).
For velocities alone, use the velocity calculator. When you know both end velocities but not the acceleration, the s = ½(u + v)t calculator is the shorter route. To find any two unknowns at once, try the SUVAT calculator.
Frequently asked questions
Why can there be two answers when I solve for time?
Solving s = ut + ½at² for t means solving a quadratic, which can have two positive roots. Physically, the object passes the same position twice: once on the way out and again after the acceleration has turned it around. The calculator shows the earlier time and gives the later one in a note.
How do I find acceleration from distance and time when starting from rest?
With u = 0 the formula becomes s = ½at², so a = 2s ÷ t². An object that covers 100 m in 5 s from rest has a = 2 × 100 ÷ 25 = 8 m/s², about 0.82 g.
When does a ball thrown straight up at 15 m/s reach a height of 10 m?
Take up as positive, so a = −9.80665 m/s². Solving for time gives 0.982 s on the way up and 2.077 s on the way down. Air resistance is ignored.
What does a negative acceleration do to the position curve?
It bends the curve downward. If the initial velocity is positive, the object slows, reaches its farthest point when v = 0 at t = −u ÷ a, and then comes back. The chart shows this as an arch instead of an upward-curving line.
Why does the note say the distance traveled is larger than |s|?
When the velocity changes sign during the interval, the object goes out and partly back. The displacement s counts only the net change in position, so the total path length is longer.