Motion in Two or Three Dimensions: Study Guide (Physics C: Mechanics 1.5)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses, and both have a Topic 1.5 about motion in two dimensions. This guide is the calculus-based one. It treats position as a vector function of time, differentiates and integrates each component, and handles an acceleration that changes with time in one direction. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Position, velocity and acceleration as vectors

In Topic 1.2 the object moved along one line, and a sign gave its direction. In a plane you need two numbers for every position. Choose axes first and write them down, for example “+x east, +y north, origin at the start” or “+x horizontal, +y up”.

The position vector points from the origin to the object:

r(t) = x(t) i + y(t) j

Here i and j are unit vectors along +x and +y. Velocity and acceleration are derivatives, exactly as before, but now of a vector:

v = dr/dt = (dx/dt) i + (dy/dt) j, so v_x = dx/dt and v_y = dy/dt

a = dv/dt = (dv_x/dt) i + (dv_y/dt) j, so a_x = dv_x/dt and a_y = dv_y/dt

Because the unit vectors i and j do not change, you differentiate (or integrate) each component on its own. Going backwards, you need an initial condition for each component:

v_x(t) = v_x0 + ∫₀ᵗ a_x dt v_y(t) = v_y0 + ∫₀ᵗ a_y dt

x(t) = x₀ + ∫₀ᵗ v_x dt y(t) = y₀ + ∫₀ᵗ v_y dt

To report a vector as a size and a direction, use Topic 1.1: speed |v| = √(v_x² + v_y²), and the angle from tan θ = v_y / v_x, checked against a sketch so you pick the right quadrant. The velocity vector is always tangent to the path. The acceleration vector need not be.

Three dimensions (background). Add a third axis: r = x i + y j + z k, with a third set of component equations. The method does not change. This course only asks you to analyse motion in two dimensions, with numbers; describing motion in three dimensions is expected in the electricity and magnetism course, not here.

One motion, two one-dimensional problems

The big idea of this topic: motion in a plane can be analysed as two separate one-dimensional motions, one along each axis. Every tool from Topics 1.2 and 1.3 applies to each component. The two components are linked only by the time, which is the same for both.

Three consequences follow.

1. The components can be different. A ball might move at 6 m/s horizontally and 2 m/s vertically. Its acceleration might be zero along x and 9.8 m/s² along y. Nothing forces the two directions to match.
2. Either component can be non-uniform. An acceleration a_x that grows with time is handled by integration, just as in one dimension, even while a_y stays constant. The constant-acceleration equations may apply along one axis and not the other.
3. Changing one component does not change the perpendicular one. Push a puck sideways on frictionless ice and its forward velocity stays the same. Throw one ball horizontally and drop another from the same height at the same moment: with air resistance ignored, they reach the floor together, because both start with v_y0 = 0 and both have a_y = −g.

A useful habit: make a two-column table, x | y, and fill in the initial position, initial velocity and acceleration for each axis before you calculate anything. Then write t only once, because it belongs to both columns.

Projectile motion: a special case

A projectile is an object moving under gravity alone near Earth’s surface, with air resistance ignored. With +x horizontal and +y up:

a_x = 0 a_y = −g

So a projectile has zero acceleration in one dimension and a constant, non-zero acceleration in the other. Launch it from the origin with speed v₀ at angle θ above the horizontal, so v_x0 = v₀ cos θ and v_y0 = v₀ sin θ. Integrate each component:

1. v_x = v₀ cos θ (constant), and x = (v₀ cos θ) t
2. v_y = v₀ sin θ − gt, and y = (v₀ sin θ) t − ½gt²

Eliminate t using t = x / (v₀ cos θ):

y = x tan θ − g x² / (2v₀² cos²θ)

This has the form y = bx − cx², so the path is a parabola. On level ground, set y = 0: the flight time is t = 2v₀ sin θ / g and the range is

R = 2v₀² sin θ cos θ / g = v₀² sin 2θ / g

Use this formula only when launch and landing heights are equal. It shows some useful patterns:

• R ∝ v₀², so doubling the launch speed multiplies the range by 4 and the flight time by 2.
• Complementary angles (for example 30° and 60°) give the same range, and 45° gives the greatest range on level ground.
• For a horizontal launch from height h, the fall time is t = √(2h/g), independent of the horizontal speed. Making h four times larger doubles the time.

Worked example 1: a stone thrown from a cliff

Question. Take +x horizontal, away from the cliff, and +y up, with the origin at sea level directly below the cliff edge. A stone is thrown from the edge, 25.0 m above the sea, at 14.0 m/s and 35.0° above the horizontal. Ignore air resistance. Find (a) the time until it reaches the sea, (b) how far from the cliff it lands, (c) its greatest height, and (d) its velocity as it enters the water.

Set up the table.

	x	y
initial position	0	25.0 m
initial velocity	14.0 cos 35.0° = 11.47 m/s	14.0 sin 35.0° = 8.03 m/s
acceleration	0	−9.8 m/s²

1. (a) Integrating a_y twice gives y = 25.0 + 8.03t − 4.9t². Set y = 0 and solve the quadratic: t = [8.03 + √(8.03² + 2 × 9.8 × 25.0)] ÷ 9.8 = 3.2 s (3.22 s). The other root is negative, so we reject it: the stone was not in flight before it was thrown.
2. (b) x = 11.47 × 3.22 = 37 m.
3. (c) At the top v_y = 8.03 − 9.8t = 0, so t = 0.82 s. Rise = 8.03² ÷ (2 × 9.8) = 3.3 m, so the greatest height is 28 m above the sea (28.3 m).
4. (d) v_x = 11.47 m/s (unchanged). v_y = 8.03 − 9.8 × 3.22 = −23.5 m/s. Speed = √(11.47² + 23.5²) = 26 m/s, at tan⁻¹(23.5 ÷ 11.47) = 64° below the horizontal.

Check. The constant-acceleration equation for y alone gives v_y² = 8.03² + 2 × 9.8 × 25.0, so |v_y| = 23.5 m/s, matching step 4.

A tempting mistake. R = v₀² sin 2θ / g gives 19 m here. That formula assumes landing at the launch height, which is not true when the stone drops 25 m below the cliff top.

Worked example 2: a rover with changing acceleration

Question. Take +x east and +y north, origin at the start, on a flat test field. A small rover starts at the origin with velocity 2.0 m/s north. Its drive gives a_x = (0.60 m/s³)t, while a steering control gives a constant a_y = −0.40 m/s². Find v(t) and r(t), the moment the rover is moving due east, and its position and speed at that moment.

1. x components (non-uniform). v_x = 0 + ∫₀ᵗ 0.60t dt = 0.30t² (m/s). x = ∫₀ᵗ 0.30t² dt = 0.10t³ (m).
2. y components (uniform). v_y = 2.0 − 0.40t (m/s). y = 2.0t − 0.20t² (m).
3. Due east means v_y = 0 while v_x > 0: 2.0 − 0.40t = 0, so t = 5.0 s.
4. At t = 5.0 s: r = (0.10 × 125) i + (10 − 5.0) j = 12.5 i + 5.0 j m. The rover is 13 m from the start, 22° north of east.
5. Speed = |v| = v_x = 0.30 × 25 = 7.5 m/s, because v_y = 0.
6. Acceleration then: a = 3.0 i − 0.40 j m/s². It is not parallel to the velocity, which shows that v and a can point in different directions.

What this shows. The two directions behave differently: v_x grows faster and faster, while v_y falls steadily. The east drive has no effect on the north motion; if it were switched off, y at 5.0 s would still be 5.0 m. Using a constant a_x = 3.0 m/s² (its value at 5.0 s) for the whole run would give v_x = 15 m/s, twice the true value.

Investigating motion in a plane

A good way to study two-dimensional motion in the lab is video analysis. Film a launched ball against a measuring scale, with the camera far away and square to the plane of motion to reduce perspective error. Step through the frames and record x and y at equal time intervals.

• Plot x against t. A straight line shows a_x = 0; its slope is v_x.
• For y, a straight-line graph is easier to judge than a curve. Since y = v_y0 t − ½gt², dividing by t gives y/t = v_y0 − (g/2)t. Plot y/t against t: the slope is −g/2 and the intercept is v_y0.
• To test that the components are independent, launch the ball horizontally at several different speeds from the same height. If the fall time stays the same while the horizontal distance grows in proportion to the speed, the evidence supports the claim.

Choose your scales so the data fill most of each axis, and label every axis with the quantity and unit.

Common misconceptions

• “The speed is zero at the top.” Only v_y is zero. The speed there is v_x0, and the acceleration is still g downward.
• “A faster horizontal throw falls more slowly.” The fall time depends only on the vertical motion (Figure 1, and the dropped-and-thrown pair).
• Using R = v₀² sin 2θ / g everywhere. It only applies when launch and landing heights are equal (Worked example 1).
• Using the constant-acceleration equations for a non-uniform component. Integrate that component instead (Worked example 2).
• Adding magnitudes of components. A velocity of 3 m/s east and 4 m/s north is 5 m/s, not 7 m/s.
• “The velocity and the acceleration point the same way.” The velocity is tangent to the path; the acceleration of a projectile always points down.
• Using a different t for each component. The time is the one thing the two directions share.

Where this leads

This is the last topic of Unit 1. In Unit 2, Force and Translational Dynamics, Newton’s second law is applied component by component, exactly like the x | y tables here, and forces that depend on time give the kind of non-uniform acceleration you met in Worked example 2. Earlier: Topic 1.4, Reference Frames and Relative Motion, where a ball thrown from a moving vehicle becomes a projectile problem in the ground frame. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.