Curved Motion & Combining Motions
Welcome to Projectile Motion. Every motion in the topics so far has run along a straight line: a car braking on a level road, a stone dropped from a height, a block pulled across a floor. A thrown ball does not move like that. It flies along a curve, and so do a falling leaf, a car on a bend and a boat that the river carries sideways. This page builds the two ideas the rest of the topic stands on. The first is what a curved path tells you about the velocity and the force: which way the velocity points, what makes a path bend, and which way it bends. The second is how one motion can be built from two simpler ones, as a boat's crossing is built from its own motion through the water and the water's motion past the banks. The next page, Horizontal Projectiles, puts the two ideas together for a ball thrown sideways.
The 2025 syllabus does not name projectiles or curved motion, yet every paper so far has tested them, with one or two questions each and six in all. Three are about directions on a curved path. January 2026 Q36 asks which velocity arrow on a falling leaf's path is drawn correctly, and April 2026 Q48 asks where the net force on a smoke particle can point. Both are worked through below. March 2026 Q23 asks the same kind of question about a thrown ball and belongs to the next page. None of the three needs any arithmetic, only two rules: the velocity lies along the path, and the force points into the bend. River crossings and boats pulled in by a rope, the usual textbook examples of combining motions, have not appeared in a paper yet, so they come last and stay short.
The velocity on a curve
The Speed & Velocity page found the direction of the velocity on a curved path. Take a point P and a later point Q on the path. The average velocity from P to Q points along the straight line PQ, and as Q is taken closer and closer to P, that line turns until, in the limit, it just touches the path at P. So the velocity at any instant points along the tangent to the path, the way the body is moving.
That has a consequence the earlier topics did not need. On a curve the tangent turns from point to point, so the direction of the velocity changes all the time, even when the speed stays the same. A velocity that changes means an acceleration, a = Δv/Δt, and on a curve it is never zero. By Newton's second law an acceleration needs a net force. So a body moving along a curve always has a net force on it: every curved motion is an accelerated motion.
The trajectory of a leaf falling from the air is shown in the figure, the velocity directions of the leaf at positions a, b, c, and d are indicated. The correct direction is
A. b B. c C. a D. d
In the figure the leaf's wavy path runs from a, at the upper left, down to the ground at the lower right, and each of the four points carries an arrow for the velocity.
The leaf is falling, so it moves along the path from a towards the ground. At each point its velocity lies along the tangent and points the way the leaf is going. Check each arrow against the path at its own point.
At a the path heads to the right and slightly upwards, but the arrow points straight down. At b the path has turned steeply downwards, but the arrow points horizontally to the right. At c the path runs to the right, but the arrow points down and a little to the left. At d the arrow points down and to the right, along the tangent, the way the leaf moves next. The answer is D.
The three wrong arrows all come from one idea: the leaf is falling, so its velocity must point down. Over the whole trip the leaf does move down, from a to the ground, but the velocity belongs to one point of the path, not to the trip. A, the arrow at b, is horizontal where the path is close to vertical. B, the arrow at c, points down where the path runs across. C, the arrow at a, points down where the path runs across and slightly up.
When a path bends
The last section showed that every curve needs a net force. The reverse is not true. A stone dropped from rest has a net force on it, its weight, and falls in a straight line. A car braking on a straight road has a net force on it, pointing backwards, and slows down without turning. What decides whether a path bends is the direction of the net force compared with the direction of the velocity.
Split the net force F into two parts, the way the Adding & Resolving Forces page split forces: one part along the velocity, Falong, and one at right angles to it, Fside. By F = ma each part gives its own part of the acceleration, and the two parts do different jobs.
Falong acts along the line of motion, like every force in the Newton's Laws topic. Pointing forwards, it speeds the body up; pointing backwards, it slows the body down. Either way it leaves the direction alone.
Fside has no part along the motion, so at that instant it does not change the speed. Over a short time Δt it gives the body a small sideways velocity, asideΔt, towards the side it points to. Added to the velocity the body already had, by the parallelogram rule, that turns the velocity a little towards that side. Repeat for the next short time, and the next, and the path curves round towards the side the force points to.
Two results follow. First, a path stays straight only if Fside is zero: the net force is zero, or it lies along the line of the velocity, pointing forwards or backwards. That covers the braking car. It also covers a body that starts from rest under a constant force: with no velocity yet, it sets off along the force and stays on that line. March 2026 Q44, worked through on the Inertia & the First Law page, is this case. A smooth ball sitting on a sliding wedge starts from rest and feels only vertical forces, so it falls straight down; option C, a parabola, was the trap.
Second, if the net force has any part at right angles to the velocity, the path bends, and it bends towards the force. So the net force, and with it the acceleration, points into the inside of the bend, the side textbooks call the concave side of the path. Seen another way, the path always lies between the direction of the velocity and the direction of the force: it starts out along v and keeps turning towards F.
What the force does to the speed
The same split tells you what happens to the speed. Only Falong changes it, and whether Falong points forwards or backwards depends on the angle between the net force and the velocity.
A ball thrown upwards at an angle shows all three. On the way up its weight points down while its velocity points up and forwards, more than 90° apart, so the ball slows. At the top the velocity is horizontal, exactly 90° from the weight. On the way down the angle is less than 90°, and the ball speeds up. The Angled Projectiles & Slopes page follows this path in detail.
To go round a bend at a steady speed, the force has to stay at exactly 90° to the velocity the whole way, turning as the velocity turns. A car on a bend at a steady speed is like this, and so is a body going round a circle at a steady speed, which the Circular Motion & Gravitation topic studies. A constant force cannot do this: it keeps its direction while the velocity turns towards it.
One bend or two: reading a curved path
A constant force has the same direction everywhere, so it always pulls the path towards the same side. The velocity keeps turning towards the direction of the force, coming closer and closer to it without ever reaching it, because the part of the velocity at right angles to the force never changes: the force has no part in that direction. So the path can turn towards the force, but it never turns back the other way. It bends one way only. A thrown ball is the standard case. Its weight is constant, its path bends downwards everywhere, and the path is a parabola. A motion with a constant acceleration is called uniformly accelerated (匀变速), so a thrown ball's motion is uniformly accelerated curvilinear motion.
Read this backwards, and a path tells you about the force. If the inside of the bend changes sides along the path, so that the path is S-shaped, the force must have changed direction between the two bends, and the motion was not uniformly accelerated. From the shape of the path alone you can also tell where the force at a point cannot point. You do not even need to know which way the body travels: the inside of a bend is on the same side whichever way you go along it.
A wind tunnel uses smoke particles to visualize airflow. A smoke particle follows a curved trajectory. Which statement is correct? ( )
A. The particle's velocity is constant.
B. The particle must move with uniformly accelerated curvilinear motion.
C. Acceleration at point P may be horizontally leftward.
D. Net force at point Q may be vertically downward.
In the figure the path runs almost level from the left, bends downwards through Q, falls to the right and levels off at P, at the bottom of the curve. No arrow shows which way the particle travels, and none is needed: the inside of each bend is the same whichever way the particle goes.
A: on a curve the direction of motion changes, so the velocity changes, whatever the speed does. False.
B: near Q the path bends downwards, so the inside of the bend is below it. Near P it bends upwards, so the inside is above it. The force points into the bend at both, so it points downwards at Q and upwards at P: it changed direction, and the acceleration was not constant. A constant force would have bent the path one way only. False.
C: at P the path is level, so the tangent there is horizontal. A horizontal acceleration lies along the tangent. It could change the speed, but it could not bend the path, and at P the path is bending upwards. The acceleration at P must point somewhere above the tangent. False.
D: at Q the path slopes down to the right and bends downwards, so the inside of the bend is below it. Straight down points into that side and is not along the tangent, which slopes, so a net force straight down can bend the path the way it bends. "May" asks for one possible direction, and this is one; down and to the left, or down and to the right, would also do. True. The answer is D.
The April paper's figure is hand-drawn, with P near the bottom of the curve. Option C is meant to be judged where the path is level, which is how P is drawn here.
Combining two motions
A boat crossing a river moves through the water, driven by its engine, while the water carries it downstream. A raindrop falls through air that the wind is carrying sideways. In each case the body takes part in two motions at once. The motion we see from the ground is called the resultant motion (合运动), and the two motions it is made of are its component motions (分运动).
The textbook demonstration uses a small wax block sealed in a long glass tube full of water. Stand the tube upright and the block rises at a steady speed, say 3 cm/s. Now, while it rises, slide the tube sideways along a bench at a steady 4 cm/s. Seen from the bench, after t seconds the block has risen 3t cm and moved 4t cm across. Its height is always ¾ of its distance across, so its positions lie on a straight line. Each second it moves 4 cm across and 3 cm up, which is √(4² + 3²) = 5 cm along that line, so it moves at a steady 5 cm/s. And the block reaches the top of the tube at the same moment whether the tube moves or not: sliding the tube does nothing to the rising.
Three facts come out of the demonstration, and each holds in general.
Whether the resultant motion is straight or curved is decided by the rule from the first half of this page, applied to the resultant velocity and the resultant acceleration. Two steady motions give no acceleration at all, so the path is a straight line, as the demonstration showed. Now let the tube speed up sideways from rest at 2 cm/s², while the block still rises at a steady 3 cm/s. The block's velocity starts out straight up and its acceleration points sideways, at right angles to it. The acceleration does not lie along the velocity, so the path curves, bending towards the side the tube accelerates to. That is the pattern of a thrown ball on the next page: a steady motion across and a uniformly accelerated motion down.
Crossing a river
A boat on a river has two velocities to combine: its velocity relative to the water, vboat, set by its engine and by the way it is pointed, and the water's velocity relative to the banks, vwater, which runs along the river. Its velocity relative to the banks is their resultant. The current runs along the river, so only vboat has a part across it, and the two motions are independent. The time to cross therefore depends on the boat alone:
t = d ÷ (the part of vboat across the river),
where d is the width of the river. While the boat crosses, the current carries it downstream.
For the shortest time, point the boat straight across, so that all of vboat goes into crossing: tmin = d/vboat, whatever the current. In that time the current carries the boat vwatertmin downstream, so it lands that far below the point opposite its start.
For the shortest path, straight across to the point opposite, the resultant velocity must point straight across, so the current has to be cancelled. Aim the boat upstream at an angle α to the bank, chosen so that the upstream part of its velocity matches the current: vboat cos α = vwater. The part left over goes across the river, vboat sin α = √(vboat² − vwater²), so this crossing takes longer than the shortest time. It works only if the boat is faster than the current. If it is not, no heading cancels the current, and the boat ends up downstream however it is pointed.
A river is 60 m wide and flows at 3 m/s. A boat moves at 5 m/s relative to the water. (a) What is the shortest time in which it can cross, and where does it land? (b) How long does it take to cross along the shortest path?
(a) Point the boat straight across: t = 60 ÷ 5 = 12 s. In those 12 s the current carries it 3 × 12 = 36 m downstream, so it lands 36 m below the point opposite its start.
(b) The boat can cancel the current, because 5 m/s is more than 3 m/s. Aim upstream with 5 cos α = 3, so cos α = 0.6 and α = 53° to the bank. The part left across the river is √(5² − 3²) = 4 m/s, and t = 60 ÷ 4 = 15 s.
Check: the shortest path takes longer than the shortest time, 15 s against 12 s, because part of the boat's velocity is spent holding it against the current.
A boat pulled in by a rope
One more resolving problem is a textbook favourite. A man on a quay pulls a boat in with a rope that runs over a pulley at the quay's edge, pulling the rope in at a known speed. How fast does the boat move?
The key is to resolve the right velocity. The boat's real motion, the resultant, is along the water surface towards the quay, at a speed v. Split that velocity into a part along the rope and a part at right angles to it. The part along the rope, v cos θ, where θ is the angle between the rope and the water, makes the stretch of rope between the boat and the pulley shorter. The part at right angles, v sin θ, carries the boat round a circle centred on the pulley for that instant, which keeps its distance from the pulley the same: it only swings the rope round. The rope is pulled in at exactly the rate at which that stretch gets shorter, so
vrope = v cos θ, and v = vrope/cos θ.
Because cos θ is less than 1, the boat moves faster than the rope. As the boat comes in, the rope gets steeper, θ grows and cos θ shrinks, so a rope pulled in at a steady speed brings the boat in faster and faster.
A boat moves towards a quay at 5 m/s, pulled by a rope over a pulley at the quay's edge. At this moment the rope makes 37° with the water. How fast is the rope being pulled in? (cos 37° = 0.8)
Resolve the boat's real velocity along the rope: vrope = v cos θ = 5 × 0.8 = 4 m/s.
The common slip resolves the rope's velocity instead, treating the boat's speed as the part of the rope's velocity along the water: v = vrope cos θ. That would make the boat slower than the rope. The boat's velocity is the real motion, the resultant, so it is the one to split.
Your turn! · Practice questions
Things to remember
| Concept | What to know |
|---|---|
| Velocity on a curve | Along the tangent, the way the body moves. Check each arrow at its own point, not against the trip as a whole. January 2026 Q36: only the arrow at d (D). |
| Curved means accelerated | The direction of v keeps changing, so a ≠ 0 and a net force acts, even at a steady speed. |
| Straight or curved | Straight if Fnet = 0, if Fnet lies along the line of v, or if the body starts from rest under a constant force (March 2026 Q44). Curved if Fnet ≠ 0 and not along v. |
| Which way it bends | Towards the force. Fnet and a point into the inside (concave side) of the bend, and the path lies between v and F. The direction of travel is not needed. |
| The speed | F less than 90° from v: speeds up. More than 90°: slows down. Exactly 90°: steady speed, and only the direction changes. |
| Constant force | Uniformly accelerated motion; the path bends one way only, and v turns towards F without reaching it. An S-shaped path means the force changed. April 2026 Q48: D. |
| May and must | "May" needs one possible case; "must" needs every case. |
| Combining motions | Component motions are independent and take the same time. Displacements, velocities and accelerations combine by the parallelogram rule; at right angles use Pythagoras, never plain addition: 3 and 4 give 5. |
| River: shortest time | Point straight across: t = d/vboat, whatever the current. The boat lands vwatert downstream. |
| River: shortest path | Only if vboat > vwater. Aim upstream with vboat cos α = vwater; then t = d/√(vboat² − vwater²), longer than the shortest time. |
| Boat on a rope | Resolve the boat's real velocity: vrope = v cos θ. The boat is faster than the rope and speeds up as θ grows. |