When acceleration is constant in a plane, the motion splits into two independent one-dimensional motions: one along x and one along y. You apply the same equations (v = u + at and s = ut + (1/2)at^2) separately to each axis, then combine the results. Memory hook: "Two straight-line motions sharing one clock" — x and y run on their own but the same time t links them.
With constant acceleration, the curved 2D path is really two independent 1D motions: apply v = u + at and s = ut + (1/2)at^2 separately to x and y, linked only by the same time t.
Your doubts, answered
How do I actually solve a 2D constant-acceleration problem?
Break every vector into x and y parts. Write the kinematics equations for x using only u_x and a_x, and separately for y using only u_y and a_y. The one thing shared by both axes is the time t. Solve for whatever you need on each axis, then recombine (for magnitude use v = sqrt(v_x^2 + v_y^2)). This is why NEET keeps the same v = u + at formula from Class 11 straight-line motion — you just use it twice.
Are the x and y motions really independent?
Yes. A constant force (constant acceleration) along one axis does not change the velocity along the perpendicular axis. That is why a ball dropped from a moving car keeps the car's horizontal speed while gravity only pulls it down. The horizontal and vertical motions do not talk to each other except through the shared time t.
Which equations do I use — is there a special vector formula?
The vector forms are v(vector) = u(vector) + a(vector) t and r(vector) = u(vector) t + (1/2) a(vector) t^2. But in NEET numericals you almost always apply the scalar versions per axis: v_x = u_x + a_x t, x = u_x t + (1/2) a_x t^2, and the same for y. Same equations, done twice.
Is projectile motion just a case of this?
Exactly. Projectile motion is motion in a plane with constant acceleration a = g downward. There a_x = 0 (uniform horizontal velocity) and a_y = -g (constant vertical acceleration). Every projectile formula comes from applying these constant-acceleration equations to each axis.
If given x(t) and y(t), how do I get acceleration?
Differentiate twice. a_x = d^2x/dt^2 and a_y = d^2y/dt^2. For example if x = 5t - 2t^2 then a_x = -4 m/s^2, and if y = 10t then a_y = 0. The total acceleration is a = sqrt(a_x^2 + a_y^2), and it is constant only if both a_x and a_y are constant (no t left in them).
⚠️ The NEET trap ✗ Students think that once an object moves in 2D, its horizontal speed must change because it is also falling. ✓ With constant acceleration in a plane, each axis is independent. If a_x = 0, the horizontal velocity u_x stays exactly the same the whole time, no matter what happens vertically. 🧠 Constant a on the y-axis cannot touch the x-axis speed. Horizontal velocity is untouched when a_x = 0.
Real NEET questions
2017
The x and y coordinates of a particle at any time t are x = 5t - 2t^2 and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is:
A · 0
B · 5 m/s^2
C · -4 m/s^2 ✓
D · -8 m/s^2
Solution: Acceleration is the second time-derivative of position, done per axis. For x: v_x = dx/dt = 5 - 4t, so a_x = dv_x/dt = -4 m/s^2 (constant, no t). For y: v_y = dy/dt = 10, so a_y = 0. Total acceleration a = sqrt(a_x^2 + a_y^2) = sqrt((-4)^2 + 0) = 4 m/s^2, directed along -x. Since a_x and a_y have no t in them, the acceleration is the same at every instant including t = 2 s. Answer: -4 m/s^2 (option C).
2021
A car starts from rest and accelerates at 5 m/s^2. At t = 4 s a ball is dropped out of a window by a person sitting in the car. What are the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s^2)
A · 20 sqrt2 m/s, 0
B · 20 sqrt2 m/s, 10 m/s^2 ✓
C · 20 m/s, 5 m/s^2
D · 20 m/s, 0
Solution: First find the ball's horizontal velocity at release. The car speeds up from rest: v_x = u + a t = 0 + 5 x 4 = 20 m/s. This is the ball's horizontal velocity at the moment it is dropped, and once free the ball has a_x = 0, so v_x stays 20 m/s. After release, gravity acts for the time from t = 4 s to t = 6 s, i.e. 2 s: v_y = g t = 10 x 2 = 20 m/s downward. Combine: speed = sqrt(v_x^2 + v_y^2) = sqrt(20^2 + 20^2) = 20 sqrt2 m/s. The ball is in free fall, so its acceleration is g = 10 m/s^2 (not zero — the car's engine no longer acts on it). Answer: 20 sqrt2 m/s, 10 m/s^2 (option B).
2019
Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings 100 m apart and of the same height 200 m, with the same speed 25 m/s. When and where will the two bullets collide? (g = 10 m/s^2)
A · after 2 s at a height of 180 m ✓
B · after 2 s at a height of 20 m
C · after 4 s at a height of 120 m
D · they will not collide
Solution: Treat horizontal (x) and vertical (y) motions separately. Horizontally, both bullets have zero vertical-independent horizontal speed 25 m/s toward each other, so they close the 100 m gap at 25 + 25 = 50 m/s. Time to meet: t = 100 / 50 = 2 s. Both start at the same height 200 m and both fall under the same g with zero initial vertical speed, so at any instant they are at the same height (their vertical motions are identical). Fall in 2 s: drop = (1/2) g t^2 = 0.5 x 10 x 2^2 = 20 m. Height of collision = 200 - 20 = 180 m. Answer: after 2 s at a height of 180 m (option A).
Solved Motion In A Plane NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It means the acceleration vector does not change in magnitude or direction over time. Because it is fixed, you can resolve it into constant a_x and a_y components and treat the x and y motions as two separate 1D constant-acceleration problems linked only by the shared time t.
What are the equations of motion for a plane with constant acceleration?
In vector form: v = u + a t and r = r_0 + u t + (1/2) a t^2. In component form (used in NEET numericals): v_x = u_x + a_x t, x = u_x t + (1/2) a_x t^2, and the same two equations for the y-axis with subscript y.
Why is horizontal velocity constant in projectile motion?
Because a_x = 0 for a projectile (gravity acts only vertically). With no horizontal acceleration, v_x = u_x stays constant for the whole flight. Only the vertical velocity changes, at the constant rate g.
How is this topic tested in NEET?
NEET asks it as short numericals: find acceleration from x(t) and y(t), find the final speed of a body dropped from a moving vehicle, or two-body collision problems using independent x and y motion. The trick is always to split into axes and share the time t.
Is total acceleration always constant if a_x and a_y are given?
Only if both a_x and a_y are themselves constant (no t remaining after differentiation). If either component still depends on time, the total acceleration changes and the simple v = u + at equations no longer apply directly.
