Motion in a plane with constant acceleration

What are the equations of motion in a plane?

motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.

NCERT Physics Part I (Class 11), Section 3.8, page 38

In plain words: a particle moving in a plane with a steady push is really doing two straight-line motions at once, one along x and one along y. Solve each with the familiar equations of motion, using the same clock for both, and then combine.

The equations in vector form

With a constant, the average acceleration over any interval equals a, so a = (v − v₀)/t (NCERT page 37):

For constant acceleration the average velocity over the interval is (v₀ + v)/2, and the displacement is the average velocity times the time (NCERT pages 37–38):

Differentiating this gives back v = v₀ + at, and at t = 0 it gives r = r₀, as it should (NCERT Eqs. 3.33a and 3.34a).

A third relation, without the time, comes from the scalar product (beyond the NCERT text). Take the scalar product of v = v₀ + at with itself:

It gives the speed, not the direction. Along a straight line it becomes v^² = v₀² + 2a(x − x₀).

Treating the x and y motions separately

In components (NCERT Eqs. 3.33b and 3.34b):

The x-equations contain only x-quantities and the y-equations only y-quantities. So:

• the x-motion is unaffected by what happens along y, and the other way round;
• the only link is the time: the particle reaches a given x and the matching y at the same instant, so find t from one equation and use it in the other;
• the speed and direction at any instant come from combining the components, v = √(v_x² + v_y²).

The same holds in three dimensions. The most important use is projectile motion under gravity, where a_x = 0 and a_y = −g (NCERT Section 3.9).

These equations need a to be constant in both size and direction. They do not apply to uniform circular motion, where the size of the acceleration is constant but its direction keeps changing (NCERT Points to Ponder 4, page 46).

If a = 0, the particle moves uniformly on a straight line or stays at rest. For non-zero constant a, the path is a line if v₀ = 0 or is parallel or antiparallel to a. Otherwise it is a parabola, as in the figure, so the same a can give either path, depending on how the motion starts (NCERT Points to Ponder 7, page 46).

Any two perpendicular axes will do (NCERT Points to Ponder 3, page 46). For a projectile in free flight near an incline of angle θ, choose x up the slope and y normal to it, away from the surface. Gravity then gives a_x = −g sin θ and a_y = −g cos θ. A block sliding in contact with a smooth fixed incline is different: its normal reaction cancels the perpendicular gravitational component, so its net a_y = 0 and a_x = −g sin θ.

Solved examples

Common mistakes to avoid

Quick revision summary

• Constant a: v = v₀ + at; r = r₀ + v₀t + ½at^².
• In components: v_x = v_0x + a_xt, x = x₀ + v_0xt + ½a_xt^², and the same for y.
• The x- and y-motions are independent one-dimensional motions linked only by the common time.
• v^² = v₀² + 2a·(r − r₀).
• Not valid when the direction of a changes, as in circular motion.

Questions students ask

Which motions have a constant acceleration vector?

Motions with a constant acceleration vector include a projectile near the Earth’s surface with air resistance neglected (a = g, vertically downwards), a block sliding freely across a smooth inclined plane (g sin θ down the line of greatest slope) and, in Class 12, a charged particle in a uniform electric field. For all of these, the equations v = v₀ + at and r = r₀ + v₀t + ½at^² hold for as long as the acceleration stays the same.

Does v^² = u^² + 2as work in two dimensions?

Yes, for constant acceleration the relation v^² = u^² + 2as holds in a plane in the vector form v^² = v₀² + 2a·(r − r₀), with a scalar product in place of as, and also separately along each axis, as v_x² = v_0x² + 2a_xΔx. It gives the speed but not the direction of motion. For example, a stone thrown at 10 m s^⁻¹ in any direction (air resistance neglected) is moving at √(100 − 2 × 9.8 × 2) ≈ 7.8 m s^⁻¹ whenever it is 2 m above the launch point, whatever the angle of the throw, provided it gets that high.

Why is the path of a particle with constant acceleration a parabola?

For non-zero constant acceleration, the path is a parabola unless the initial velocity is zero or along the acceleration line. If the acceleration itself is zero, the motion is a straight line at constant velocity (or rest). Take the y-axis along a and start at the origin: across the acceleration the motion is uniform, x = v_0xt, and along it uniformly accelerated, y = v_0yt + ½at^². Eliminating t = x/v_0x gives y = (v_0y/v_0x)x + ax^²/(2v_0x²), a quadratic in x and so a parabola whenever v_0x ≠ 0; projectile motion, with a = g, is the best-known case.

Can a particle with constant acceleration in a plane come to rest?

A particle with constant acceleration comes to rest, even for an instant, only if its velocity is zero to begin with or points exactly opposite to the acceleration, so that it moves along a straight line. Otherwise the part of the velocity perpendicular to a never changes, and the speed can never fall below it. For example, with v₀ = 10 m s^⁻¹ at 120° to an acceleration of 2 m s^⁻², the 5 m s^⁻¹ component against a is cancelled after 5/2 = 2.5 s, when the speed reaches its minimum, 10 sin 120° ≈ 8.7 m s^⁻¹; after that the speed rises again.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. v₀ = (1î + 2ĵ) m/s and a = (2î − 1ĵ) m s^⁻². Find the velocity at t = 4 s.

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   v = (1 + 8)î + (2 − 4)ĵ = (9î − 2ĵ) m/s.

2. Q2. Starting from rest at the origin with a = (4î + 3ĵ) m s^⁻², how far from the origin is a particle after 2 s?

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   r = ½at^² = ½(4î + 3ĵ) × 2^² = (8î + 6ĵ) m, so it is √(8^² + 6^²) = 10 m from the origin.

3. Q3. Why can the x- and y-motions be solved separately?

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   In the component equations each axis has only its own quantities: x-motion depends on v_0x and a_x, y-motion on v_0y and a_y. The two share only the time.