Kinematics Equations
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Direct answer
Three equations govern motion with constant acceleration: v = u + at, s = ut + (1/2)at^2 and v^2 = u^2 + 2as, all obtainable from the v–t graph, where the slope gives acceleration and the area under the curve gives displacement. Two companions matter as much in JEE: the displacement in the nth second, s(n) = u + (a/2)(2n − 1), which is a difference of positions, not a position; and the average velocity (u + v)/2, valid only for uniform acceleration. When acceleration varies with time or position, these closed forms die and you must integrate, using a = v dv/dx when acceleration is given as a function of velocity and position.
What you must remember
- Core set: v = u + at; s = ut + (1/2)at^2; v^2 = u^2 + 2as; each uses one sign convention — fix the positive direction before writing anything.
- Displacement in the nth second: s(n) = u + (a/2)(2n − 1); it has the dimensions of velocity times one second, and it is never the position of the particle.
- Average velocity: (u + v)/2 only for constant a; in general, average velocity = total displacement/total time, and average speed = total distance/total time — equal only for motion that never reverses.
- Graph reading: slope of x–t is velocity, slope of v–t is acceleration, area under v–t is displacement, area under a–t is change in velocity; a curved x–t graph means changing velocity.
- Free fall symmetry: time up equals time down, u(up) = v(down) at the same level, and total flight time from level to level is 2u/g.
- Variable acceleration toolkit: v = dx/dt, a = dv/dt = v dv/dx; the last identity skips time and is the standard route when a is a function of x.
- Relative shortcut: two bodies under identical acceleration have zero relative acceleration, so separation changes at a constant rate.
- Pattern note: Main tests these equations as direct numericals; Advanced rarely devotes a full question to them but buries them inside collisions, rotating frames and constrained systems.
A meeting problem solved the smart way
A ball is dropped from a height h at the same moment another is thrown upward from directly below with speed u. The brute-force method writes two quadratic positions and subtracts; the intelligent method notices that both accelerations equal g downward, so their relative acceleration is zero and the gap h closes at the constant rate u. They meet after t = h/u. The meeting height follows from the lower ball: y = u t − (1/2) g t^2 = h − g h^2/(2u^2).
The same discipline governs the nth-second trap. A body starts from rest with a = 2 m/s^2: distance in the 3rd second is (2/2)(2 × 3 − 1) = 5 m, while distance in 3 seconds is 9 m. Questions deliberately ask the first while dressing themselves as the second, and the 4 m gap between those answers is exactly the mark Main is hunting for.
Where students slip
Three faults dominate. Mixing signs — taking g positive for one part of flight and negative for another without redrawing the axis — corrupts an otherwise correct solution; choose upward or downward positive once per problem. Misreading the nth-second formula as a position produces answers off by whole metres, the single most common Kinematics error in Main. And applying v^2 = u^2 + 2as across a journey where acceleration changes (a lift speeds up then moves uniformly) is invalid; the equation holds per segment, and the piecewise v–t graph — whose total area must equal the stated displacement — is the honest bookkeeper. Graph questions punish the same carelessness: a negative v–t portion is backward motion, and its area subtracts from displacement while adding to distance.
Frequently asked questions
When is the average velocity formula (u + v)/2 valid?
Only for constant acceleration; otherwise compute total displacement divided by total time, or integrate the velocity over the interval.
How does displacement in the nth second differ from displacement in n seconds?
s(n) = u + (a/2)(2n − 1) gives the distance covered only during the nth second, whereas s = un + (1/2)an^2 accumulates all n seconds — the two coincide only trivially at n = 1.
Why is the v–t graph so central to kinematics?
Its slope is the acceleration and its area is the displacement, so any piecewise motion — however complicated — can be read off one diagram without memorising extra equations.
What is a = v dv/dx used for?
It eliminates time when acceleration depends on position or velocity, turning v dv = a dx into a direct integral between initial and final speeds.
Can the three equations of motion be used for a particle with changing acceleration?
No — they assume uniform acceleration; split the motion into constant-acceleration segments or switch to calculus.