The Displacement Calculator (v, a, t) finds net change in position from initial velocity, acceleration and elapsed time, using the SUVAT equation that includes a squared time term. Enter u, a and t, and the tool returns displacement with the substitution shown, without needing to know the final velocity at all.
This is the form to use for free-fall and launch problems, where a starting speed and a constant acceleration (often gravity) are known and the question asks how far the object travels in a given time.
Calculate displacement from initial velocity, acceleration and time
Initial velocity, acceleration and time together fix displacement through an equation that includes a term in t². Final velocity is not required.
s = ut + ½at²Worked example: a cart starts at u = 5 m/s, accelerates at a = 2 m/s² for t = 3 s.
1. Calculate the linear term. u t = 5 × 3 = 15.
2. Calculate the squared term. ½ a t² = ½ × 2 × 3² = ½ × 2 × 9 = 9.
3. Add the two terms. 15 + 9 = 24 m.
Apply the formula to a free-fall problem
An object dropped from rest falls under standard gravity, g = 9.80665 m/s². With u = 0, the equation simplifies since the first term drops out entirely. Worked example: find the distance fallen after 3 seconds, starting from rest. **1.
List what is known.** u = 0, a = 9.80665 m/s², t = 3 s.
2. Apply the equation. s = (0 × 3) + ½ × 9.80665 × 3².
3. Substitute. s = 0 + ½ × 9.80665 × 9 = ½ × 88.25985.
4. Solve. s = 44.13 m (rounded to four significant figures).
Air resistance is ignored in this idealized calculation, and local gravity varies slightly with latitude and altitude; coursework generally treats 9.80665 m/s² (or the rounded 9.81) as constant.
Rearrange for the other variables
The same equation can be solved for initial velocity or acceleration when displacement, time, and one of those two are known instead.
u = (s − ½at²)/t
a = 2(s − ut)/t²Solving for time instead requires the quadratic formula, since t appears both linearly and squared in the original equation; that case is covered on the full Displacement Calculator.
Understand why the squared term appears
The s = ut + ½at² equation comes from integrating velocity over time. The ut term is what displacement would be if velocity stayed constant at its initial value. The ½at² term is the extra distance covered because velocity keeps increasing (or decreasing) throughout the interval due to acceleration.
Doubling the time doesn't just double this extra term, it quadruples it, since time is squared, which is why falling objects cover much more distance in each successive second than the one before.
Watch for negative acceleration
When acceleration opposes the initial velocity, such as a ball thrown upward with gravity acting downward, the ½at² term subtracts rather than adds. Displacement can even become negative if the object has enough time to return past its starting point.
Keeping a consistent sign convention for u, a and the resulting s throughout the whole problem prevents this case from producing a confusing or wrong-looking answer.
Work through a deceleration example
A car moving at u = 20 m/s brakes at a = −4 m/s² for t = 2 s, before it would come to a stop. 1. Calculate the linear term. u t = 20 × 2 = 40. **2.
Calculate the squared term.** ½ a t² = ½ × (−4) × 2² = ½ × (−4) × 4 = −8.
3. Add the two terms. 40 + (−8) = 32 m.
The negative acceleration term reduces the total displacement below the 40 m the car would have covered at a constant 20 m/s, since the braking force works against the direction of motion throughout the 2-second interval.
Compare displacement over unequal time intervals
Because time is squared in the ½at² term, displacement does not grow at a constant rate even under constant acceleration. For the free-fall example above, doubling the time from 3 to 6 seconds gives s = ½ × 9.80665 × 36 ≈ 176.52 m, four times the 44.13 m covered in the first 3 seconds rather than double.
That fourfold growth for a doubled time interval is the direct signature of the squared term in the equation.
Frequently asked questions
What is the formula for displacement with initial velocity, acceleration and time?
s = ut + ½at². For u = 5 m/s, a = 2 m/s² and t = 3 s, s = 15 + 9 = 24 m.
How far does something fall in 3 seconds from rest?
Using g = 9.80665 m/s², a 3-second fall from rest covers s = ½ × 9.80665 × 9 ≈ 44.13 m.
Do I need final velocity to use this formula?
No. This equation uses only initial velocity, acceleration and time; final velocity is not required and can be found separately if needed.
What if acceleration is negative?
A negative acceleration subtracts from the ut term. If it's large enough relative to time, the total displacement can itself become negative, meaning the object ends up behind its starting point.
Can this formula solve for time instead of displacement?
Yes, but solving for time requires the quadratic formula because t appears as both t and t² in the equation. The full Displacement Calculator covers that case with both roots explained.
Is g = 9.8 or 9.80665 used for free fall?
Standard gravity is defined as exactly 9.80665 m/s². Many textbooks round this to 9.81 or 9.8; this tool uses the exact defined value for consistency.
Does this equation work for horizontal motion too?
Yes. The equation applies to any constant-acceleration motion along a single axis, whether that axis is vertical (as in free fall) or horizontal (as in a car accelerating from a stoplight).
Summary
Displacement from initial velocity, acceleration and time follows s = ut + ½at², giving 24 m for u = 5 m/s, a = 2 m/s² and t = 3 s, and about 44.13 m for a 3-second free fall from rest at standard gravity.
The squared time term accounts for the extra distance covered as velocity keeps changing throughout the interval, which is why this form is the standard choice for free-fall and launch problems where final velocity isn't given.