Kinematics Equations (u, v, a, t, s) — Choosing the Right One
Most students who lose marks in kinematics do not lose them because the algebra is hard. They lose them because they picked the wrong equation, then spent four minutes solving a quadratic they never needed. There is a simple trick that removes the guessing completely, and this article teaches it: look at which variable the question never mentions. That one missing variable tells you exactly which equation to write down.
The five variables
Every uniform-acceleration problem in Class 9 to Class 12 physics — and in JEE, NEET and GATE — is built from the same five quantities:
| Symbol | Meaning | SI unit |
|---|---|---|
| u | Initial velocity (velocity at the start of the interval) | m/s |
| v | Final velocity (velocity at the end of the interval) | m/s |
| a | Acceleration — must be constant for these equations | m/s² |
| t | Time taken for the interval | s |
| s | Displacement (not distance — direction matters) | m |
Because the letters are s, u, v, a, t, these are often called the SUVAT equations. They apply only when acceleration is constant. If acceleration changes with time, you need calculus instead.
The five equations
2. s = ut + ½at² (no v)
3. v² = u² + 2as (no t)
4. s = ½(u + v)t (no a)
5. s = vt − ½at² (no u)
The missing-variable trick
Each equation contains exactly four of the five variables. So one variable is always absent. Now read the question and make a small list: what am I given, and what am I asked for? That is four quantities. The fifth one — the one the question neither gives you nor asks for — is the variable you want to be absent. Pick the equation that leaves it out, and you solve in one step.
| Variable not involved | Use this equation |
|---|---|
| s (displacement) | v = u + at |
| v (final velocity) | s = ut + ½at² |
| t (time) | v² = u² + 2as |
| a (acceleration) | s = ½(u + v)t |
| u (initial velocity) | s = vt − ½at² |
Worked example 1 — a car starting from rest
Question: A car starts from rest and accelerates uniformly at 2 m/s² for 6 s. Find its final velocity and the distance covered.
List: u = 0, a = 2 m/s², t = 6 s. Asked: v, then s.
For v, displacement s is not involved → use equation 1:
v = u + at = 0 + (2)(6) = 12 m/s
For s, we now know u, a and t, and v is not needed → equation 2:
s = ut + ½at² = (0)(6) + ½(2)(6²) = 0 + ½(2)(36) = 36 m
Check with equation 4: s = ½(u + v)t = ½(0 + 12)(6) = (6)(6) = 36 m. ✔
Worked example 2 — braking, where time is missing
Question: A car moving at 20 m/s brakes uniformly and stops after travelling 50 m. Find the acceleration, then the time taken.
List: u = 20 m/s, v = 0, s = 50 m. Asked: a. Time t is not given and
not asked → use equation 3:
v² = u² + 2as
0 = (20)² + 2a(50)
0 = 400 + 100a
100a = −400 → a = −4 m/s²
The minus sign is not a mistake. It says the acceleration points opposite to the motion — that is what braking means. Never drop it.
Now for t, displacement is no longer needed → equation 1:
v = u + at → 0 = 20 + (−4)t → 4t = 20 → t = 5 s
Check with equation 4: s = ½(20 + 0)(5) = (10)(5) = 50 m. ✔
Worked example 3 — a ball thrown straight up
Question: A ball is thrown vertically upward at 19.6 m/s. Taking g = 9.8 m/s², find the time to reach the highest point and the maximum height.
Take upward as positive. Then u = +19.6 m/s and a = −9.8 m/s², because gravity pulls downward. At the highest point the ball is momentarily at rest, so v = 0.
Time (s not needed) → equation 1:
0 = 19.6 + (−9.8)t → 9.8t = 19.6 → t = 2 s
Maximum height (t not needed) → equation 3:
0 = (19.6)² + 2(−9.8)s
0 = 384.16 − 19.6s → 19.6s = 384.16 → s = 19.6 m
Check with equation 2: s = (19.6)(2) + ½(−9.8)(2²) = 39.2 − 19.6 = 19.6 m. ✔
Worked example 4 — when the initial velocity is missing
Equation 5 is the one students forget exists, so examiners like using it.
Question: A train decelerates uniformly at 1.5 m/s². After 8 s its velocity is 10 m/s. How far did it travel in those 8 s?
List: v = 10 m/s, a = −1.5 m/s² (deceleration), t = 8 s. Asked: s.
The initial velocity u is neither given nor asked → equation 5:
s = vt − ½at²
s = (10)(8) − ½(−1.5)(8²)
s = 80 − ½(−1.5)(64)
s = 80 − (−48) = 128 m
Check by finding u first: u = v − at = 10 − (−1.5)(8) = 10 + 12 = 22 m/s. Then s = ½(22 + 10)(8) = (16)(8) = 128 m. ✔
Common mistakes that cost marks
- Mixing up sign conventions mid-question. Choose one positive direction at the start, write it down, and keep it for the whole problem. If up is positive, then g = −9.8 m/s² everywhere in that solution.
- Using these equations when acceleration is not constant. If a question says "acceleration a = 3t", SUVAT does not apply at all.
- Treating displacement as distance. A ball thrown up that returns to your hand has s = 0, but the distance travelled is twice the maximum height.
- Forgetting to square the time. In s = ut + ½at², it is t², not t. This single slip is the most common arithmetic error in board answer sheets.
- Not converting units. 72 km/h is 72 × 1000 ÷ 3600 = 20 m/s. Mixing km/h with m/s² gives an answer that is wrong by a factor of 3.6.
- Dropping the negative root. Equation 3 gives v², so v could be + or −. Decide from the physics which sign is real.
Where kinematics appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 9–10 | Motion in a straight line, distance–time and velocity–time graphs |
| CBSE/ICSE Class 11 | Motion in a straight line, free fall, relative velocity |
| JEE Main & Advanced | Projectile motion, two-body chase problems, graph-based questions |
| NEET | One-mark direct substitution questions — speed matters more than depth |
| GATE (engineering papers) | Rigid-body and particle dynamics starting from these equations |
One last tip: the graph shortcut
On a velocity–time graph with constant acceleration, the line is straight. The slope of that line is the acceleration and the area under it is the displacement. Equation 4, s = ½(u + v)t, is just the area of a trapezium with parallel sides u and v and width t. If you ever forget the equations under exam pressure, sketch the graph and read them off.
Check your working in seconds. Enter any three of u, v, a, t, s and the Kinematics tool solves for the rest, keeping the signs straight so you can see whether your negative acceleration was handled correctly.
Open the Kinematics Calculator →Studying physics and chemistry for Class 11–12 boards? ABC Chemistry runs Class 11–12 coaching at the Gurugram centre and online classes across India, with home tuition available in the Delhi-NCR area — details at abcchemistry.in.