A particle returns to its starting point after moving along a closed path. Which is true?

Average velocity is zero but average speed is not

Both average speed and average velocity are zero

Average velocity is zero but average speed is not

Average speed is zero but average velocity is not

A bullet is fired horizontally from a height and another is simply dropped from the same height at the same instant. They:

The fired bullet lands later, having travelled further

It depends on the mass of each bullet

Rain falls vertically at 10 m/s. A man walks east at 5 m/s. He should hold his umbrella:

Tilted forward (eastward) at about 27° from the vertical

Vertically, since the rain falls vertically

Tilted forward (eastward) at about 27° from the vertical

Tilted backward (westward) at about 27° from the vertical

Tilted forward at 45° from the vertical

A boat can move at 5 m/s in still water and must cross a 100 m wide river flowing at 3 m/s. To reach the point directly opposite, it should be aimed:

Upstream at an angle θ with sin θ = 3/5

Straight across, perpendicular to the bank

Upstream at an angle θ with sin θ = 3/5

Downstream at an angle θ with sin θ = 3/5

In a velocity–time graph, the area between the curve and the time axis represents:

Displacement, with area below the axis counted as negative

Kinematics — Straight Line, Plane Motion, Projectiles and Relative Velocity

Kinematics is where most students first meet the difference between knowing a formula and knowing when it applies. The three equations of motion are memorised in a day; the mistakes that cost marks come from applying them to non-uniform acceleration, from treating distance and displacement as the same thing, and from failing to see that a two-dimensional problem is just two one-dimensional problems that share a clock. This chapter is built around that last idea, because once you can separate the axes, projectiles and relative motion stop being separate topics and become arithmetic.

Motion in a Straight Line

The distinctions that questions are built on
Pair	Distance / speed (scalar)	Displacement / velocity (vector)
Definition	Total path length; never decreases	Change in position; can be zero or negative
Magnitude relation	distance ≥ \|displacement\|, equal only for straight-line motion without reversal	Average velocity = displacement / time
Round trip	Average speed is non-zero	Average velocity is exactly zero

For uniform acceleration only, the three equations are v = u + at, s = ut + ½at², and v² = u² + 2as, together with s in the nth second = u + (a/2)(2n − 1). Every one of these fails the moment acceleration varies with time or position, and that failure is a favourite question.

⚠️ When acceleration is not constant, the formulae are useless

If a question gives you a = kt, or a = kv, or a = kx, you must go back to calculus. The three standard forms are a = dv/dt when acceleration is a function of time, a = v·dv/dx when it is a function of position, and for a = f(v) you separate as dv/f(v) = dt. The middle one is the one students forget, and it comes from the chain rule: a = dv/dt = (dv/dx)(dx/dt) = v·dv/dx. Recognising which form to use takes five seconds and is the entire difficulty of these questions — a candidate who reaches for v² = u² + 2as when a depends on x has already lost the mark.

🧠 Read graphs by slope and area

Two rules answer every graph question. Slope: the slope of a position–time graph is velocity, and the slope of a velocity–time graph is acceleration. Area: the area under a velocity–time graph is displacement, and the area under an acceleration–time graph is the change in velocity. Two refinements matter. Area below the axis counts as negative, so for distance travelled you add magnitudes while for displacement you keep signs — that single distinction is the most common graph question in the paper. And a position–time graph can never be vertical, because that would mean infinite velocity; questions offer such a graph as an option precisely to see whether you notice.

Projectile Motion — Two Independent Problems

The whole subject rests on one observation: the horizontal and vertical motions are independent and share only the time. Horizontally there is no acceleration, so x = (u cos θ)·t. Vertically there is a constant −g, so the vertical motion is exactly the straight-line problem you already solved. Every projectile result follows from those two lines.

Projectile from ground level, launched at θ with speed u
Quantity	Result	Worth noticing
Time of flight	T = 2u sin θ / g	Twice the time to the top — the path is symmetric
Maximum height	H = u² sin²θ / 2g	At the top the vertical velocity is 0, not the whole velocity
Range	R = u² sin 2θ / g	Maximum at θ = 45°, where R = u²/g
Velocity at any instant	vₓ = u cos θ always; v_y = u sin θ − gt	The horizontal component never changes
Path	A parabola: y = x tan θ − gx²/(2u²cos²θ)	Eliminating t from the two equations gives it

🧠 Two angles, one range — and the fact that follows

Because sin 2θ = sin(180° − 2θ), the angles θ and (90° − θ) give the same range. So 30° and 60° land in the same place, as do 20° and 70°. The higher angle takes longer and goes higher; the lower one is flatter and faster. Two exam consequences follow. First, a question saying "two projectiles have the same range" is telling you the angles are complementary, which is usually the key step. Second, for those two angles T₁T₂ = 2R/g and H₁ + H₂ = u²/2g — compact results that turn a two-projectile question into one line.

⚠️ Horizontal projection is not the same formula

A body thrown horizontally from a height h has initial vertical velocity zero, so its time of fall is t = √(2h/g) — the same as if it were simply dropped. That is the classic result: a bullet fired horizontally and a bullet dropped from the same height hit the ground together, because the horizontal velocity does not affect the vertical motion at all. Its horizontal range is u√(2h/g). Students routinely misapply the ground-level T = 2u sin θ/g here, which gives zero because sin 0° = 0 — and a zero time of flight for an object that visibly does fly should be caught immediately as absurd.

Relative Velocity — Rain, Rivers and Closest Approach

One definition does all the work: v⃗_AB = v⃗_A − v⃗_B, the velocity of A as seen by B. Everything else is vector subtraction, and the reason students find this topic hard is almost always that they are trying to picture it instead of subtracting components.

The three standard situations
Situation	Set-up	Result
River crossing — shortest time	Point the boat straight across, perpendicular to the bank	t = d/v_boat, but you drift downstream by (v_river · t)
River crossing — shortest path	Aim upstream at angle θ with sin θ = v_river/v_boat	Resultant is straight across at √(v_b² − v_r²); impossible if v_river > v_boat
Rain and an umbrella	v⃗_rain,man = v⃗_rain − v⃗_man	Tilt the umbrella along the relative velocity; tan θ = v_man/v_rain from the vertical

🎯 Why rain appears to slant forward when you run

Rain falling vertically at 10 m/s while you walk east at 5 m/s has, relative to you, a velocity of 10 m/s down and 5 m/s west — because subtracting your eastward velocity adds a westward component. So the rain appears to come from in front, and you must tilt the umbrella forward by tan⁻¹(5/10) ≈ 27° from the vertical. This is worth understanding rather than memorising, because the sign is where marks are lost: students tilt the umbrella backwards, reasoning that they are moving forwards. The rule that never fails is to compute v⃗_rain − v⃗_man as components and point the umbrella along the answer.

🧠 Closest approach — sit on one of the bodies

Two particles moving on different straight lines: when are they nearest? Doing this in the ground frame means minimising a messy function of t. Instead put yourself on particle B. Now B is at rest and A moves in a straight line with constant velocity v⃗_AB, so the shortest distance is simply the perpendicular distance from B to that line — a geometry problem with no calculus at all. The same trick makes collision questions trivial: two bodies collide if and only if v⃗_AB points along the line joining them initially, which is far easier to check than solving two position equations simultaneously.

Key takeaways

Distance ≥ |displacement| always; on a round trip average velocity is exactly zero but average speed is not.
The three equations of motion hold ONLY for uniform acceleration.
Use a = dv/dt when a depends on t, and a = v·dv/dx when a depends on x.
Slope of x–t is v, slope of v–t is a; area under v–t is displacement, under a–t is Δv.
Projectile: T = 2u sinθ/g, H = u²sin²θ/2g, R = u²sin2θ/g, maximum range at 45°.
θ and (90° − θ) give the same range; horizontal velocity u cosθ never changes during flight.
Thrown horizontally from height h: t = √(2h/g), the same as being dropped.
v⃗_AB = v⃗_A − v⃗_B. For closest approach, sit on one body and take the perpendicular distance.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A particle returns to its starting point after moving along a closed path. Which is true?
1. Both average speed and average velocity are zero
2. Average velocity is zero but average speed is not
3. Average speed is zero but average velocity is not
4. Neither can be determined
Show answer
Answer: B — Average velocity is zero but average speed is not
Average velocity is displacement over time, and for a closed path the displacement is exactly zero. Average speed is total path length over time, which is positive because the particle did travel. This asymmetry between the scalar and the vector is the single most tested idea in one-dimensional kinematics.
The acceleration of a particle is given by a = kx, where x is its position. To find v as a function of x you should use:
1. v² = u² + 2ax
2. a = dv/dt
3. a = v·dv/dx
4. v = u + at
Show answer
Answer: C — a = v·dv/dx
Since a is a function of position, write a = v·dv/dx (from the chain rule, dv/dt = (dv/dx)(dx/dt)) and separate as v dv = kx dx. The equation v² = u² + 2as is invalid here because it assumes constant acceleration, and reaching for it is the standard error in this question type.
A projectile is launched at 60° to the horizontal. Another with the same speed achieves the same range when launched at:
1. 15°
2. 30°
3. 45°
4. 75°
Show answer
Answer: B — 30°
Range is u² sin 2θ/g, and sin 120° = sin 60°, so θ and (90° − θ) give the same range — here 60° and 30°. The 30° projectile is flatter and lands sooner; the 60° one goes higher and takes longer. A question stating that two projectiles share a range is usually telling you the angles are complementary.
A bullet is fired horizontally from a height and another is simply dropped from the same height at the same instant. They:
1. Reach the ground together
2. The fired bullet lands later, having travelled further
3. The dropped bullet lands later
4. It depends on the mass of each bullet
Show answer
Answer: A — Reach the ground together
The horizontal and vertical motions are independent, and both bullets have zero initial vertical velocity, so both take t = √(2h/g). The horizontal velocity changes where the fired bullet lands, not when. Mass is irrelevant because g is the same for all masses in free fall.
At the highest point of a projectile's path (launched at 60°, speed u), the speed of the projectile is:
1. Zero
2. u/2
3. u√3/2
4. u
Show answer
Answer: B — u/2
At the top the vertical component is zero but the horizontal component u cos 60° = u/2 is unchanged, so the speed is u/2 rather than zero. Choosing zero is the classic error — it confuses "the vertical velocity vanishes" with "the velocity vanishes", and u√3/2 comes from using sin 60° instead of cos 60°.
Rain falls vertically at 10 m/s. A man walks east at 5 m/s. He should hold his umbrella:
1. Vertically, since the rain falls vertically
2. Tilted forward (eastward) at about 27° from the vertical
3. Tilted backward (westward) at about 27° from the vertical
4. Tilted forward at 45° from the vertical
Show answer
Answer: B — Tilted forward (eastward) at about 27° from the vertical
Relative to the man, v⃗_rain − v⃗_man is 10 m/s down and 5 m/s west, so the rain appears to come from in front and the umbrella tilts forward by tan⁻¹(5/10) ≈ 27°. Tilting backward is the standard error, from reasoning about one's own motion instead of computing the relative velocity; 45° would require the two speeds to be equal.
A boat can move at 5 m/s in still water and must cross a 100 m wide river flowing at 3 m/s. To reach the point directly opposite, it should be aimed:
1. Straight across, perpendicular to the bank
2. Upstream at an angle θ with sin θ = 3/5
3. Downstream at an angle θ with sin θ = 3/5
4. Upstream at 45°
Show answer
Answer: B — Upstream at an angle θ with sin θ = 3/5
To land directly opposite, the upstream component of the boat's velocity must cancel the current, so v_boat sin θ = v_river gives sin θ = 3/5. The resultant across is √(5² − 3²) = 4 m/s, taking 25 s. Aiming straight across instead gives the shortest time but drifts downstream — the two questions have different answers, so read which is asked.
In a velocity–time graph, the area between the curve and the time axis represents:
1. Acceleration
2. Displacement, with area below the axis counted as negative
3. Distance, always positive
4. The change in acceleration
Show answer
Answer: B — Displacement, with area below the axis counted as negative
The area under a v–t graph is displacement and carries sign, so area below the time axis is negative. For distance travelled you would instead add the magnitudes of the areas. Acceleration is the slope of this graph, not its area — confusing slope with area is the other common error here.