Sometimes you know how fast an object was moving at the start and at the end of an interval, and how long the interval lasted, but not the acceleration. A speedometer reading at each end of a highway on-ramp is a typical case. If the velocity changed at a steady rate, the displacement is simply the average of the two velocities times the elapsed time. This calculator applies that rule, solves it for any one of its four quantities, and draws the velocity–time graph whose shaded area is the answer.
How to use the s = ½(u + v)t calculator
- Under Solve for, choose Displacement, Initial velocity, Final velocity or Time.
- Enter the other three values: Displacement (s), Initial velocity (u), Final velocity (v) and Time (t), each with its own unit.
- Use Show the result in to pick the output unit.
- Alongside the answer you get the average velocity ½(u + v), the implied acceleration (v − u) ÷ t and the velocity–time diagram.
The s = ½(u + v)t formula
Rearranged for the other three unknowns:
v = 2s ÷ t − u
t = 2s ÷ (u + v)
Velocities and displacement are signed along the direction you call positive. Time must be positive. When you solve for time, the calculator rejects three situations: u + v = 0 (the average velocity is zero), s = 0, and a displacement whose sign is opposite to that of ½(u + v), since no positive time can fit.
Worked example: merging onto a highway
Given: a car enters the ramp at 10 m/s and reaches 30 m/s after 8 s.
Average velocity: ½ × (10 + 30) = 20 m/s
s = 20 m/s × 8 s = 160 m
Implied acceleration: (30 − 10) ÷ 8 = 2.5 m/s²
The same formula works for slowing down. A train braking from 25 m/s to 5 m/s over 600 m takes t = 2 × 600 ÷ (25 + 5) = 40 s, and its implied acceleration is −0.5 m/s².
Reading the velocity–time diagram
The diagram plots velocity on the vertical axis against time. With constant acceleration the velocity is a straight line from u to v, and the region between that line and the time axis is a trapezoid. Its area, the average of the two parallel sides times the width, is exactly ½(u + v)t. That is the formula in geometric form, and it is why the shaded region equals the displacement.
The area idea also explains direction reversals. Take u = 20 m/s, v = −10 m/s and t = 6 s. The implied acceleration is −5 m/s², so the object stops at t = 4 s and then moves backward.
| Part of the motion | Time span | Area | Meaning |
|---|---|---|---|
| Above the axis | 0 to 4 s | ½ × 20 × 4 = +40 m | 40 m forward |
| Below the axis | 4 to 6 s | ½ × (−10) × 2 = −10 m | 10 m back |
| Net | 0 to 6 s | ½ × (20 − 10) × 6 = 30 m | displacement |
Areas below the axis count as negative, so the net displacement is 30 m, while the distance actually traveled is 50 m. The calculator returns the 30 m, shades the region below the axis in red, and notes that u and v have opposite signs, with the turnaround at 4 s and the 50 m distance.
Why the midpoint rule needs constant acceleration
Averaging the two end velocities is valid only when the velocity changes linearly with time. Then the velocity spends as much time above the midpoint as below it. If the acceleration varies, the velocity–time curve bows above or below the straight line, and ½(u + v) no longer equals the true average.
Here is an example of how far off it can be. A car goes from rest to 20 m/s in 4 s, then cruises at 20 m/s for another 6 s. Over the whole 10 s its true displacement is 40 m + 120 m = 160 m. Plugging u = 0, v = 20 m/s and t = 10 s into the formula gives only 100 m, because it assumes the car took the full 10 s to reach 20 m/s. To get the right answer, split the motion into stretches of constant acceleration and add them, or enter the stretches as legs in the average velocity calculator.
If you know the acceleration rather than the final velocity, use the s = ut + ½at² calculator. To find a missing end velocity from the acceleration and time first, use the velocity calculator. For acceleration from a change in velocity, or from force and mass, see the acceleration calculator.
Frequently asked questions
Do I need to know the acceleration to use s = ½(u + v)t?
No. The formula only needs the two end velocities and the time, which is its main advantage. The acceleration still has to be constant, and the calculator works it out for you as (v − u) ÷ t.
Can I enter velocities in km/h?
Yes. Each velocity field has its own unit menu. A car going from 36 km/h to 72 km/h in 10 s averages 15 m/s, so it covers 150 m, with an implied acceleration of 1 m/s².
Why can't the calculator find the time when u + v = 0?
If the initial and final velocities are equal and opposite, the average velocity is zero. Any displacement divided by zero is undefined, so the time cannot be recovered from this formula. Use s = ut + ½at² with a known acceleration instead.
What happens if the velocity changes sign during the interval?
The formula still gives the correct net displacement, because the part of the motion in the negative direction subtracts. It does not give the total distance traveled, which is larger. The calculator adds a note with the distance traveled whenever u and v have opposite signs, and the graph shades the backward part in red.