Equations of Motion in Two Dimensions (Vector Form)
Physics · Motion In A Plane · NEET
In two dimensions with constant acceleration, the same equations as 1D still work, but written as vectors: v = u + a t and r = r0 + u t + (1/2) a t^2. The trick for NEET is to split each vector into x and y components and solve the x-motion and y-motion separately, because the two directions do not affect each other. Memory hook: "One vector equation = two separate scalar equations (x and y run in parallel lanes)."
A curved 2D path with the net velocity vector v split into independent x and y components; the same two vector equations govern both lanes, then components recombine via speed = sqrt(vx^2 + vy^2).
Your doubts, answered
Are the equations of motion different in 2D compared to 1D?
No. The formulas are the same, just written as vectors: v = u + a t and r = r0 + u t + (1/2) a t^2. In 1D these are single numbers; in 2D each is a vector with an x-part and a y-part. So one vector equation quietly contains two scalar equations. This is why you never need to memorise 'new' 2D formulas for NEET.
Why do we split motion into x and y components?
Because the x-direction and y-direction are independent. A force or acceleration along x only changes the x-velocity; it does nothing to the y-velocity, and the reverse is also true. So you apply v = u + at once for x (using ux, ax) and once for y (using uy, ay). This 'divide and conquer' step is the single most tested idea in projectile and 2D kinematics questions.
How do I find velocity and acceleration from a position vector r(t)?
Differentiate. If r(t) = x(t) i + y(t) j, then velocity v = dr/dt = (dx/dt) i + (dy/dt) j, and acceleration a = dv/dt = (d^2x/dt^2) i + (d^2y/dt^2) j. Differentiate each component on its own. Example: r = 4t i + 2t^2 j gives v = 4 i + 4t j, so at t = 1 s, v = 4 i + 4 j.
Does motion along x affect motion along y?
No, they are completely separate for constant acceleration. A ball thrown horizontally keeps the same horizontal speed while gravity acts only on the vertical part. That is why a bullet fired horizontally and a bullet dropped from the same height hit the ground at the same time: their vertical motions are identical and independent of any horizontal motion.
What is the vector form of v = u + at?
It is v = u + a t, where v, u and a are all vectors. Written in components: vx = ux + ax t and vy = uy + ay t. You solve these two lines separately, then combine them at the end using magnitude = sqrt(vx^2 + vy^2) and direction = tan^-1(vy / vx).
⚠️ The NEET trap ✗ Students see two directions and add the numbers directly (e.g. treat a horizontal 5 m/s^2 and vertical 10 m/s^2 as one 15 m/s^2), or find velocity by adding speeds arithmetically. ✓ Keep x and y separate. Compute vx and vy on their own with vx = ux + ax t and vy = uy + ay t, then combine with magnitude = sqrt(vx^2 + vy^2). Two perpendicular parts of 20 m/s each give 20*sqrt(2), NOT 40 m/s. 🧠 Adding accelerations before resolving into components
Real NEET questions
NEET 2017
The x and y coordinates of a particle at any time t are x = 5t - 2t^2 and y = 10t, 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: Step 1: Find ax by differentiating x twice. dx/dt = 5 - 4t, so d^2x/dt^2 = -4 m/s^2. Step 2: Find ay by differentiating y twice. dy/dt = 10, so d^2y/dt^2 = 0. Step 3: The acceleration vector is a = -4 i + 0 j, which is constant, so it is the same at every instant including t = 2 s. Magnitude = 4 m/s^2 directed along -x, i.e. -4 m/s^2. Answer: -4 m/s^2 (option C).
NEET 2023
The position of a particle is given by r(t) = 4t i + 2t^2 j + 5 k (t in seconds, r in metres). Find the magnitude and direction (with respect to the x-axis) of the velocity v(t) at t = 1 s.
A · 3*sqrt(2) m/s, 30 degrees
B · 3*sqrt(2) m/s, 45 degrees
C · 4*sqrt(2) m/s, 45 degrees ✓
D · 4*sqrt(2) m/s, 60 degrees
Solution: Step 1: Velocity is the time derivative of the position vector, component by component. v = dr/dt = 4 i + 4t j + 0 k. Step 2: Put t = 1 s: v = 4 i + 4 j. Step 3: Magnitude = sqrt(4^2 + 4^2) = sqrt(32) = 4*sqrt(2) m/s. Step 4: Direction with x-axis = tan^-1(vy / vx) = tan^-1(4/4) = tan^-1(1) = 45 degrees. Answer: 4*sqrt(2) m/s at 45 degrees (option C).
NEET 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*sqrt(2) m/s, 0
B · 20*sqrt(2) m/s, 10 m/s^2 ✓
C · 20 m/s, 5 m/s^2
D · 20 m/s, 0
Solution: Step 1 (horizontal, x): When released at t = 4 s, the car's speed is vx = ux + ax t = 0 + 5*4 = 20 m/s. The ball keeps this horizontal velocity (no horizontal force after release), so vx stays 20 m/s. Step 2 (vertical, y): After release the ball falls freely for 6 - 4 = 2 s. vy = g*t = 10*2 = 20 m/s. Step 3: Combine the perpendicular components: speed = sqrt(vx^2 + vy^2) = sqrt(20^2 + 20^2) = 20*sqrt(2) m/s. Step 4: Once released, the only acceleration is gravity, a = g = 10 m/s^2. Answer: 20*sqrt(2) m/s, 10 m/s^2 (option B).
Solved Motion In A Plane NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What are the equations of motion in two dimensions in vector form?
v = u + a t and r = r0 + u t + (1/2) a t^2, where u, v, a and r are vectors. Each splits into an x-equation and a y-equation that you solve separately.
Can I use v = u + at directly in 2D?
Yes, as long as acceleration is constant. Apply it once to the x-components (vx = ux + ax t) and once to the y-components (vy = uy + ay t), then combine the results.
How do I get the final speed and direction from components?
Speed = sqrt(vx^2 + vy^2) and direction from the x-axis = tan^-1(vy / vx). Never add vx and vy arithmetically unless they point along the same line.
Why is this concept important for NEET?
Almost every projectile, river-boat, and r(t) question in Motion in a Plane reduces to applying these two vector equations component-wise. Mastering the split-into-components step lets you solve a whole family of NEET problems with one method.
How do I get velocity and acceleration if position is given as a function of time?
Differentiate the position vector once for velocity and twice for acceleration, treating each component (x, y, z) separately.
