The Displacement Calculator (u, v, t) finds net change in position from initial velocity, final velocity and elapsed time, without needing acceleration as a separate input. Enter u, v and t, and the tool returns displacement using the average of the two velocities multiplied by time, with the arithmetic shown.
This is the form to reach for when both end velocities are known but acceleration is not directly given, which is a common setup in problems that describe how fast something started and how fast it ended, without stating the rate of change explicitly.
Calculate displacement from initial and final velocity
When both end velocities and the time are known, displacement is the average of those velocities times the interval. Acceleration drops out of the formula entirely, which is why this form works even when the acceleration value is unknown or irrelevant to the question being asked.
s = ½(u + v)tWorked example: u = 5 m/s, v = 11 m/s, t = 3 s.
1. Add the two velocities. 5 + 11 = 16.
2. Multiply by time. 16 × 3 = 48.
3. Divide by two. 48 / 2 = 24 m.
Rearrange for the other variables
The same equation can be rearranged to solve for any of the four quantities when the other three are known.
u = 2s/t − v
v = 2s/t − u
t = 2s/(u + v)Worked check: solve for time given s = 24 m, u = 5 m/s, v = 11 m/s.
1. Add the velocities. 5 + 11 = 16.
2. Apply the rearranged formula. t = 2 × 24 / 16.
3. Solve. t = 48 / 16 = 3 s, confirming the original example.
Know when the rearrangement fails
If u + v equals zero and s is not zero, time cannot be found from the rearranged formula, because dividing by zero is undefined.
Physically, this means the object has equal and opposite end velocities, for example moving forward at 5 m/s and finishing at −5 m/s, a case that needs a different equation, one that keeps acceleration explicit, to solve for time correctly.
Understand why acceleration is not needed
This equation works because it captures the average velocity directly from the two endpoints, rather than deriving that average from acceleration and time.
As long as acceleration is constant throughout the interval, the average of the initial and final velocity always equals the true average velocity over that interval, and average velocity times time always equals displacement, regardless of how that average arose.
That is what makes the formula valid without ever mentioning acceleration.
Compare with the acceleration-based form
The equivalent displacement formula that does use acceleration is s = ut + ½at². Both formulas return the same displacement for the same physical scenario; they simply start from different known quantities. If a problem gives u, a and t but not v, the acceleration-based form is more direct.
If a problem gives u, v and t but not a, this form skips a step by not requiring acceleration to be calculated first.
Work through a second worked example
An object starts at u = 2 m/s and finishes at v = 14 m/s over t = 4 s. 1. Add the two velocities. 2 + 14 = 16. 2. Multiply by time. 16 × 4 = 64. **3.
Divide by two. 64 / 2 = 32 m**.
The same three-step process, add, multiply by time, then halve, produces a correct result regardless of how large the two velocities are, as long as both are measured along the same axis with a consistent sign convention.
Check the result against average velocity directly
Another way to confirm the 24 m result is to compute the average velocity first, then multiply separately: (5 + 11) / 2 = 8 m/s average, and 8 × 3 = 24 m, the same answer found through the combined formula above.
Splitting the calculation into these two steps can make the reasoning easier to follow, even though the single-line formula produces the identical result in one pass.
Frequently asked questions
What is the formula for displacement from initial and final velocity?
s = ½(u + v)t, the average of the two velocities multiplied by time. For u = 5 m/s, v = 11 m/s and t = 3 s, s = ½ × 16 × 3 = 24 m.
Do I need acceleration to use this formula?
No. This form calculates displacement directly from the two end velocities and time, without needing acceleration as a separate input.
Can I solve this formula for time instead of displacement?
Yes: t = 2s/(u + v). This works as long as u + v is not zero.
What happens if u and v are equal and opposite?
If u + v = 0, the rearranged formula for time is undefined, since it requires dividing by zero. A different equation involving acceleration is needed to solve for time in that case.
Is this the same as s = vt with average velocity?
Yes. Substituting v̄ = (u + v)/2 into s = v̄t produces exactly this same formula, just written with u and v spelled out rather than pre-averaged.
Does this formula assume constant acceleration?
Yes. The averaging relationship between initial and final velocity holds specifically because acceleration is constant across the interval; with varying acceleration, the true average velocity would need to be found differently.
What if I know u, a and t but not v?
Use s = ut + ½at² instead, which is the direct form for that combination of known values, available on the main Displacement Calculator.
Summary
Displacement from initial velocity, final velocity and time follows s = ½(u + v)t, giving 24 m for u = 5 m/s, v = 11 m/s and t = 3 s. This form skips acceleration entirely, relying instead on the fact that average velocity under constant acceleration is simply the mean of the two endpoints.
The same equation rearranges cleanly for u, v or t, except when u and v are equal and opposite, in which case a different SUVAT equation is needed to solve for time.