Motion in a Plane: Position, Velocity and Acceleration Vectors
Physics · Motion In A Plane · NEET
In a plane, motion is described by three vectors: position r = x i + y j (where the particle is), velocity v = dr/dt (how fast and which way it moves), and acceleration a = dv/dt (how velocity changes). Velocity is always along the path (tangent), while acceleration can point in any direction and bends the path. Memory hook: "r tells WHERE, v tells d(WHERE)/dt, a tells d(v)/dt" — each vector is just the time-rate of the one before it.
At point P on a curved path, the position vector r points from the origin O to P, the velocity v is tangent to the path, and the acceleration a can point in any direction (here inward, bending the path). v = dr/dt and a = dv/dt.
Your doubts, answered
How do I get velocity and acceleration from a position vector r(t)?
Differentiate with respect to time. If r = 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 separately. Example: r = 4t i + 2t^2 j gives v = 4 i + 4t j and a = 4 j (constant). This is the single most tested skill from this concept in NEET.
Is velocity always tangent to the path?
Yes. In two dimensions the velocity vector always points along the direction of motion, which is the tangent to the trajectory at that instant. This is because v = dr/dt points in the direction the position is instantaneously changing. Acceleration, however, is NOT restricted to the tangent — it can be along, opposite, or perpendicular to velocity.
Why can acceleration point sideways even when speed is constant?
Acceleration is the rate of change of the velocity VECTOR, not just its magnitude. If the direction of velocity changes (like in circular motion) while the speed stays constant, the velocity vector still changes, so a is non-zero. That acceleration points perpendicular to velocity, toward the centre. This is why uniform circular motion is accelerated motion.
What is the difference between speed and the magnitude of velocity?
For instantaneous values they are the same: speed = |v|. The trap is average quantities. Average speed = total path length / time, while magnitude of average velocity = |displacement| / time. Since path length is greater than or equal to displacement, average speed is greater than or equal to |average velocity|.
Can velocity be perpendicular to the position vector?
Yes. This happens whenever v . r = 0. The classic case is uniform circular motion about the origin, r = cos(wt) i + sin(wt) j, where v = w(-sin(wt) i + cos(wt) j) and v . r = 0 always. Here velocity is tangent to the circle while r is the radius, so they are perpendicular, and acceleration a = -w^2 r points to the centre.
⚠️ The NEET trap ✗ Treating average velocity magnitude the same as average speed, or assuming acceleration must be along velocity. ✓ Velocity magnitude equals speed only instantaneously; averaged, |avg velocity| = displacement/time can be far less than avg speed = path/time. Acceleration is d(velocity vector)/dt, so it can be perpendicular to v even at constant speed. 🧠 Constant speed does NOT mean zero acceleration. If the DIRECTION of v changes, a is non-zero.
Real NEET questions
NEET 2023
The position of a particle is given by r(t) = 4t i + 2t^2 j + 5 k, where t is in seconds and 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: Differentiate the position vector. v = dr/dt = d/dt(4t) i + d/dt(2t^2) j + d/dt(5) k = 4 i + 4t j + 0. Step 2: Put t = 1 s. v = 4 i + 4 j m/s. Step 3: Magnitude |v| = sqrt(4^2 + 4^2) = sqrt(32) = 4 sqrt(2) m/s. Step 4: Direction from x-axis, theta = tan^-1(v_y / v_x) = tan^-1(4/4) = tan^-1(1) = 45 degrees. Answer: 4 sqrt(2) m/s at 45 degrees (option C).
NEET 2016
A particle moves so that its position vector is given by r = cos(wt) x-hat + sin(wt) y-hat, where w is a constant. Which of the following is true?
A · Velocity and acceleration both are perpendicular to r
B · Velocity and acceleration both are parallel to r
C · Velocity is perpendicular to r and acceleration is directed towards the origin ✓
D · Velocity is perpendicular to r and acceleration is directed away from the origin
Solution: Step 1: Velocity v = dr/dt = w(-sin(wt) x-hat + cos(wt) y-hat). Step 2: Check v . r = w[(-sin(wt))(cos(wt)) + (cos(wt))(sin(wt))] = 0, so velocity is perpendicular to r. Step 3: Acceleration a = dv/dt = w^2(-cos(wt) x-hat - sin(wt) y-hat) = -w^2 r. The minus sign means a points opposite to r, i.e. towards the origin (centripetal). Answer: option C. This is uniform circular motion of radius 1 about the origin.
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 three key vectors in motion in a plane?
Position vector r = x i + y j, velocity vector v = dr/dt, and acceleration vector a = dv/dt = d^2r/dt^2. Each is found by differentiating the previous one with respect to time, component by component.
How is displacement different from position vector?
Position vector r locates the particle from the origin at one instant. Displacement is the change in position, delta r = r_final - r_initial, a vector from the start point to the end point. Velocity is defined using displacement: average v = delta r / delta t.
Does uniform circular motion have acceleration?
Yes. Even though the speed is constant, the direction of velocity keeps changing, so the velocity vector changes and a = -w^2 r is non-zero. It points toward the centre and is called centripetal acceleration.
How do I find the angle of the velocity vector?
Use theta = tan^-1(v_y / v_x), measured from the positive x-axis. Check the signs of v_x and v_y to place the angle in the correct quadrant.
Why is this concept important for NEET?
It is the foundation for projectile motion, circular motion and relative velocity, which together give several questions each year. NEET directly asks you to differentiate a given r(t) to get v and a, so mastering it secures easy marks.
