Physics I: Mechanics › Kinematics › full formula sheet

Δx = ½(v₀ + v)tSay it: displacement equals one half the sum of initial and final velocity, times time

The average-velocity form

The kinematic equation that never asks about acceleration — perfect for the problems where a is missing.

Δx is displacement, v₀ initial velocity, v final velocity, t time. Valid only when acceleration is constant (velocity must rise in a straight line).

Before this lesson: Displacement

Where it comes from

Picture a car cruising at 10 m/s that smoothly accelerates to 30 m/s over 5 s. How far did it go? You could find the acceleration first (4 m/s²) and then grind through Δx = v₀t + ½at² — but that drags in a variable the question never asked about. There is a shorter road: because the velocity rose steadily, the car’s average pace was exactly the midpoint of 10 and 30.

Before reading on: the car’s speed climbs steadily 10→30 m/s over 5 s. What was its average velocity during those 5 s — and how far did it travel? Try it before the formula confirms you.
v̄
=
(10 + 30)/2 = 20 m/s
Steady rise → the average sits exactly halfway between the endpoints.
Δx
=
v̄t = 20 · 5 = 100 m
Displacement is average velocity times time — no acceleration needed anywhere.

That two-step move is the formula: Δx = ½(v₀ + v)t. It is the right tool whenever a problem hands you the two velocities and the time but stays silent about acceleration.

Derivation

Two routes to the same equation — the algebra route and the geometry route. Both take one line once you see the midpoint idea.

Δx
=
v₀t + ½at²
Step 1 (algebra route) — start from the displacement formula we just derived. Now eliminate a using a = (v − v₀)/t.
=
v₀t + ½·[(v − v₀)/t]·t²
Step 2 — substitute. One t cancels: ½·(v − v₀)·t.
=
v₀t + ½(v − v₀)t = ½(v₀ + v)t
Step 3 — collect. v₀t − ½v₀t = ½v₀t, leaving the midpoint form. ∎

Geometry route: on a v–t graph with constant acceleration, the area under the curve is a trapezoid with parallel sides v₀ and v and width t. Area of a trapezoid = ½(sum of parallel sides)·width = ½(v₀ + v)t. And area under v–t is displacement. Same equation, zero algebra.

How to use it

The procedure, every time:

  1. Check the givens. This form shines when you know v₀, v, and t but not a. If a is given and t is missing, reach for the time-free equation instead.
  2. Verify constant acceleration. “Accelerates steadily / uniformly / at a constant rate” — or a is simply constant. If the push varies, the midpoint is wrong.
  3. Average the velocities: (v₀ + v)/2, then multiply by t.
  4. Read the answer’s sign. Negative Δx means net motion toward the negative direction.

The rearranged forms

t = 2Δx / (v₀ + v)solve for the time — “how long did the trip take?”
v = 2Δx/t − v₀solve for the final velocity — “how fast was it going at the end?”Say it: twice the displacement over time, minus where it started

Example: Δx = 200 m, v₀ = 10 m/s, v = 40 m/s → t = 2·200/50 = 8 s. No acceleration computed, none needed.

Common mistake: using the midpoint average when acceleration varied — e.g., a car with air drag, or a rocket burning fuel. The (v₀+v)/2 shortcut is a constant-acceleration privilege; revoke it the moment a changes.
Common mistake: averaging the velocities but forgetting to multiply by t — reporting “20 m/s” as the displacement. The average is a velocity; Δx needs the ×t.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: 5→25 m/s in 10 s

  1. List givens. v₀ = 5 m/s, v = 25 m/s, t = 10 s. Acceleration not given, not needed.
  2. Midpoint: (5 + 25)/2 = 15 m/s.
  3. Multiply by t: Δx = 15 · 10 = 150 m.
  4. Cross-check with the other form. a = (25−5)/10 = 2 m/s²; Δx = 5·10 + ½·2·100 = 50 + 100 = 150 m ✓
Common mistake: computing (25 − 5)/2 = 10 m/s — averaging the change instead of the values. The midpoint is (v₀ + v)/2, a sum, not a difference.
Your turn — 2→18 m/s in 8 s. Displacement?

Answer: 80 m. (2 + 18)/2 = 10 m/s; 10 · 8 = 80 m.

Example 2 — braking: 24→0 m/s in 6 s

  1. List givens. v₀ = 24 m/s, v = 0 m/s, t = 6 s.
  2. Midpoint: (24 + 0)/2 = 12 m/s — the car averaged half its initial speed while braking steadily.
  3. Multiply: Δx = 12 · 6 = 72 m of stopping distance.
  4. Sense-check. Unbraked it would travel 24·6 = 144 m; steady braking to a stop covers exactly half ✓.
Common mistake: writing Δx = 24 · 6 = 144 m — using the initial velocity as if it held the whole time. Braking means the velocity fell; the average is what counts.
Your turn — braking 16→0 m/s in 4 s. Stopping distance?

Answer: 32 m. (16 + 0)/2 = 8 m/s; 8 · 4 = 32 m.

Example 3 — solve for time: Δx = 200 m, 10→40 m/s

  1. Rearrange first: t = 2Δx/(v₀ + v).
  2. Plug in. t = 2·200/(10 + 40) = 400/50.
  3. Compute. = 8 s.
  4. Verify. Midpoint velocity 25 m/s · 8 s = 200 m ✓.
Common mistake: t = Δx/(v₀ + v) — dropping the 2. Then 200/50 = 4 s, and the check fails: 25 · 4 = 100 ≠ 200. The 2 comes from the ½ in the original — always carry it.
Your turn — Δx = 150 m, 5→25 m/s. Time?

Answer: 10 s. t = 2 · 150/30 = 300/30 = 10 s.

Before reading on: Δx = 90 m in t = 6 s, starting at v₀ = 12 m/s. The average velocity is 15 m/s — so what must the final velocity be? Reason it out before the algebra.

Example 4 — solve for v: Δx = 90 m, v₀ = 12 m/s, t = 6 s

  1. Start from the form. 90 = ½(12 + v)·6 = 3(12 + v).
  2. Divide. 30 = 12 + v.
  3. Solve. v = 18 m/s.
  4. Verify. Midpoint (12+18)/2 = 15 m/s; 15·6 = 90 m ✓.
Common mistake: solving 90 = (12 + v)·6 — forgetting the ½. That gives v = 3 m/s, which is slower than the start while covering 90 m in 6 s (average 15 m/s) — impossible, since the average of 12 and 3 is 7.5, not 15.
Your turn — Δx = 140 m, v₀ = 10 m/s, t = 8 s. Final velocity?

Answer: 25 m/s. 140 = ½(10 + v) · 8 = 4(10 + v); 35 = 10 + v; v = 25 m/s.

Memorization tips

  • Say it aloud: “displacement is the average of the two velocities, times the time.” Midpoint, then multiply.
  • Trapezoid picture: on a v–t graph the area is a trapezoid with sides v₀ and v. ½(v₀+v)t is that area — draw it once and the formula is unforgettable.
  • No-a detector: problem gives v₀, v, t but no a? This is your equation. Problem gives a but no t? Use the time-free one.
  • The 2 travels with the ½: every rearranged form (t = 2Δx/(v₀+v), v = 2Δx/t − v₀) carries a 2. Lose it and the answer halves.
  • Constant-a only: the midpoint is exact for steady acceleration and wrong otherwise. “Steadily” in the problem statement is your green light.
  • Check via midpoint: after solving, compute (v₀+v)/2 · t and confirm it equals Δx. Ten seconds, catches sign slips.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and Average velocity form is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

When do I use delta-x = 1/2 (v0 + v) t?

When you know the initial velocity, final velocity, and time — but not the acceleration. It is the only kinematic equation with no 'a' in it, so it is the fastest route for exactly those problems.

Why is the average velocity (v0 + v)/2?

Under constant acceleration, velocity rises in a straight line on a v-t graph, so its average over the interval sits exactly halfway between the endpoint values. This is the midpoint rule — and it fails the moment acceleration varies.

How is this related to the area under a v-t graph?

Directly: displacement is the area under the velocity-time curve. With constant acceleration that area is a trapezoid with parallel sides v0 and v and width t, whose area is 1/2(v0 + v)t — the same formula.

Can I use it if acceleration is not constant?

No. The (v0+v)/2 midpoint is only exact for constant acceleration. With varying acceleration (drag, throttles, springs) the true average velocity differs, and this form gives the wrong displacement.

How do I solve for time with this equation?

Rearrange to t = 2 delta-x / (v0 + v). The 2 comes from clearing the 1/2 — dropping it is the most common algebra slip on this form.

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