Displacement Calculator
Calculate displacement, initial velocity, acceleration, or time using the kinematic equation s = ut + ½at². Select what you want to find, enter the known values, and get an instant answer.
Displacement Calculator Physics Kinematics
Displacement Formula
Where:
- \(s\) — displacement of the object (m)
- \(u\) — initial velocity (m/s)
- \(a\) — constant acceleration (m/s²)
- \(t\) — time elapsed (s)
This is one of the fundamental kinematic equations of physics. It gives the position change of an object that moves with constant acceleration \(a\), starting with velocity \(u\), over a time interval \(t\). The equation rearranges to solve for any of the other variables:
\[ u = \frac{s - \frac{1}{2}at^2}{t}, \quad a = \frac{2(s - ut)}{t^2}, \quad t = \frac{-u + \sqrt{u^2 + 2as}}{a} \]
Displacement – Calculation Example
Example: a stone is dropped from rest (\(u = 0\) m/s) and falls with an acceleration of \(9.8\) m/s² for \(5\) seconds. How far does it fall?
\[ s = ut + \frac{1}{2}at^2 = 0 \times 5 + \frac{1}{2} \times 9.8 \times 5^2 \]
\[ s = 0 + 4.9 \times 25 = 122.5 \text{ m} \]
Result: the stone travels \(122.5\) m in 5 seconds — the same answer the calculator returns with its default values.
How to Use the Calculator
- In the "I want to calculate" dropdown, choose the unknown quantity: Displacement (s), Initial Velocity (u), Acceleration (a), or Time (t).
- Enter the Initial Velocity (u) and pick its unit — m/s or km/h.
- Enter the Acceleration (a) in m/s².
- Enter the Time (t) and choose seconds or minutes.
- If you are solving for u, a, or t, also enter the known Displacement (s) in meters or kilometers.
- Click Calculate. The result appears instantly, and the field you are solving for is hidden since its value is not needed as input.
What Is Displacement?
Displacement is the change in position of an object: the straight-line distance from its starting point to its ending point, together with the direction. It differs from distance travelled, which counts the total path length. If you walk 3 m east and then 3 m west, you covered a distance of 6 m, but your displacement is zero because you ended where you started.
The formula \(s = ut + \frac{1}{2}at^2\) comes from the kinematics of uniformly accelerated motion. The first term, \(ut\), is the distance the object would cover at its initial velocity alone. The second term, \(\frac{1}{2}at^2\), is the extra distance gained (or lost) because the velocity keeps changing at rate \(a\). This equation is one of the standard set of kinematic equations, alongside \(v = u + at\) and \(v^2 = u^2 + 2as\).
Interpreting the result is straightforward: a positive displacement means the object moved in the direction you treat as positive; a negative value means it moved the other way. Note that the calculator always assumes motion along a straight line. For motion in two dimensions, such as a ball thrown at an angle, use the Projectile Motion Calculator instead.
Assumptions and Limitations
- Constant acceleration only. The formula is exact only when acceleration is uniform. If acceleration changes over time (e.g. air drag becoming significant), the result is an approximation.
- Straight-line motion. Only the component of motion along one axis is handled; horizontal and vertical motion must be computed separately.
- Solving for time. When you solve for \(t\), the calculator uses \(t = \frac{-u + \sqrt{u^2 + 2as}}{a}\) and returns the earliest positive solution. If \(u^2 + 2as < 0\) or \(a = 0\) with nonzero displacement and zero velocity, no valid solution exists and an error is shown.
- Units are converted automatically. Velocity entered in km/h is converted to m/s, time in minutes to seconds, and displacement in km to meters before computing, so you can mix units freely.
Practical Tips
A common source of confusion is the sign of acceleration. Gravity near Earth's surface gives \(a \approx 9.8\) m/s² downward. If you define upward as positive, use \(a = -9.8\) m/s² for a falling object; if downward is positive, use \(+9.8\) m/s². The default values in this calculator (\(u = 0\), \(a = 9.8\), \(t = 5\)) describe an object in free fall from rest.
This equation also applies to braking problems: enter a negative acceleration for a decelerating vehicle and the displacement becomes its stopping distance. To average results over a whole trip rather than analyzing a single acceleration phase, see the Average Speed Calculator.
Frequently Asked Questions
How do I calculate displacement by hand without the calculator?
Multiply initial velocity by time, then add half of acceleration times time squared: \(s = ut + \frac{1}{2}at^2\). For example, with \(u = 5\) m/s, \(a = 2\) m/s², and \(t = 3\) s, you get \(s = 15 + 9 = 24\) m. Keep all values in SI units (meters, seconds) before you start, and watch the signs of \(u\) and \(a\) — they must use the same positive direction.
What is the difference between displacement and distance travelled?
Displacement is the straight-line change in position with direction, while distance is the total path length. They are equal only when the object moves in one direction without reversing. A ball thrown up and caught at the same height travels a long distance but has zero displacement. This calculator always returns displacement, so if the object turns around, the answer can be smaller than the actual path covered.
Why did I get a negative displacement?
A negative result simply means the object ended up opposite to the direction you defined as positive. If you call upward positive and enter \(a = -9.8\) m/s² for a falling object, a negative displacement means it moved down — which is physically correct. Check the signs of your inputs rather than the size of the answer: mixing sign conventions (e.g. positive velocity with negative acceleration inconsistently) is the most common mistake.
Can I use this formula when acceleration is zero?
Yes. With \(a = 0\) the formula reduces to \(s = ut\), ordinary constant-velocity motion. You can safely enter 0 for acceleration when solving for displacement, initial velocity, or time. Only solving for acceleration becomes impossible in that case, and solving for time with the quadratic formula needs special handling — if the object has constant velocity, just divide displacement by velocity instead.
What happens if I enter velocity in km/h or time in minutes?
The calculator converts everything to SI units automatically: km/h is divided by 3.6 to get m/s, minutes are multiplied by 60 to get seconds, and kilometers are multiplied by 1000 to get meters. You can mix units freely — for example, velocity in km/h with time in seconds. The result is shown in the unit you selected for the target quantity.
How is this equation related to v = u + at and v² = u² + 2as?
All three are the standard kinematic equations for constant acceleration, derived from calculus. Use \(v = u + at\) when you need final velocity, \(v^2 = u^2 + 2as\) when time is unknown, and \(s = ut + \frac{1}{2}at^2\) when final velocity is unknown. Any problem solvable with one can be solved with the others; pick the equation whose variables match what you already know.
Does this work for a car braking to a stop?
Yes, as long as braking is roughly uniform. Enter the car's speed as initial velocity, use a negative acceleration for deceleration, and the displacement is the stopping distance. For example, at \(u = 20\) m/s with \(a = -5\) m/s², the car stops in 40 m. Real braking distances vary with road conditions, tires, and reaction time, so treat the result as an idealized estimate, not a safety figure.