Describing Motion Around Us
In this chapter we focus on two of these — linear motion (motion in a straight line) and uniform circular motion. You already know distance, time and speed from earlier classes. Now you will learn three new and important physical quantities:
- Displacement — how far and in which direction an object has moved from its start.
- Average velocity — how fast the position changes, with direction.
- Average acceleration — how fast the velocity itself changes.
You will also learn to describe motion in three ways: in words, using numbers and equations, and using graphs. Master all three and no motion problem can surprise you in the exam.
When an object moves along a straight line, its motion is called linear motion or motion in a straight line. This is the simplest kind of motion. You see it all around you — a swimmer in a straight lane, a ball falling vertically, a car on a straight highway, or a train on a straight track. To study any motion, the very first thing we must be able to do is describe where the object is — its position — at different moments of time.
To say where an object is, we must first fix a starting point to measure from. This fixed point is called the reference point (or origin). We then state the object’s distance and direction from this reference point. Both are needed — distance alone is not enough, because the object could be on either side.
The distance and direction of an object with respect to a reference point, at a given instant of time, describes the position of the object at that instant.
Once we can describe position, we can decide whether an object is moving or still:
- If the position of the object changes with time (relative to the reference point), the object is in motion.
- If the position does not change with time, the object is at rest.
Because a straight-line object can move only two ways — forward or backward — we use + and − signs for direction. By convention, positions to the right of the origin O are positive, and positions to the left are negative. The diagram below shows an athlete’s reference line, with her start marked as origin O and two later positions B and A:
Now imagine the athlete actually runs. She starts at O (t = 0 s), reaches B at t = 4 s, runs on to A at t = 10 s, then turns back and returns to B at t = 16 s. Let us measure her journey in two different ways.
First, the total distance travelled — the entire length of path her feet actually covered. She went O→A (100 m) and then back A→B (60 m), so:
Total distance = OA + AB = 100 + 60 = 160 m
Second, we ask a different question: how far is her final position from her starting position? She ended at B, which is only 40 m from O. This “start-to-end” quantity is called displacement.
Displacement is the net change in the position of an object between two given instants of time. It has both magnitude and direction.
A few key points that examiners love to test:
- The magnitude of a quantity is its numerical value with units, without direction.
- Displacement needs both a number and a direction (e.g. “40 m in the positive direction”). Distance needs only a number.
- The SI unit of both distance and displacement is the metre (m).
- In our example, distance (160 m) and displacement (40 m) are clearly not equal — this happens whenever the object turns back.
Scalars are quantities specified by numerical value only — e.g. distance, speed. Vectors need both magnitude and direction — e.g. displacement, velocity, acceleration.
| Position | Total distance from O | Displacement from O |
|---|---|---|
| O | 0 cm | 0 cm |
| A | 40 cm | 40 cm up |
| B | 140 cm | 140 cm up |
| C | 140 + 60 = 200 cm | 80 cm up |
| O | 140 + 140 = 280 cm | 0 cm |
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Displacement is the net change in the position of an object between two instants of time. It is a vector (it has both magnitude and direction).Show Answer
Total distance = 100 + 100 = 200 m. Displacement = 0 (she returns to the start, so net change in position is zero).Show Answer
Fuel depends on the total distance travelled, not displacement. The engine works over every metre of the actual path. A car that drives out and returns home has zero displacement but has still burned fuel for the whole trip — proving fuel tracks distance, not displacement.We can now describe an object’s position and how much it has moved. But there is still a question we haven’t answered: how fast is it moving? Two quantities answer this — average speed and average velocity. They sound similar but are not the same, and knowing the exact difference is worth easy marks.
Let us start with the one you already know. Average speed tells us how fast an object covers ground, without caring about direction. It is simply the total distance divided by the time taken:
The average speed of an object is the total distance travelled divided by the time interval during which the distance is covered. It is a scalar. SI unit: metre per second (m s⁻¹).
Because distance has no direction, average speed also has no direction — only a numerical value. Speed also lets us classify motion into two types:
- Uniform motion — the object covers equal distances in equal intervals of time. Its speed stays constant.
- Non-uniform motion — the object covers unequal distances in equal intervals of time. Its speed keeps changing (increasing, decreasing, or both).
Speed tells us how fast, but not which way. In many real situations we also need the direction — and that is where average velocity comes in. Velocity uses displacement (which has direction) instead of distance:
The average velocity of an object in a time interval is the change in position (displacement) divided by the time interval in which the change occurs. It is a vector. SI unit: metre per second (m s⁻¹).
Since velocity comes from displacement, it carries the same direction as the displacement, shown by a + or − sign. Both average speed and average velocity share the same SI unit, m s⁻¹ (also written m/s), and are commonly measured in km h⁻¹ on the road. Velocity is the rate of change of position — the ratio of change in position to the time taken.
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Total distance = 400 km, total time = 5 h → average speed = 400 ÷ 5 = 80 km h⁻¹. Displacement = 200 − 200 = 0 → average velocity = 0 km h⁻¹.Show Answer
(i) When the object moves in a straight line in one direction only (no turning back) — then distance = magnitude of displacement, so the two are equal. (ii) When the object returns to its starting point — displacement is zero (velocity zero), but the path covered is not zero (speed not zero).Velocity itself can change — a bus speeds up from a stop, or slows down at a signal. When you feel a “jolt” as a vehicle starts or stops suddenly, you are feeling a change in velocity. The quantity that measures how quickly velocity changes is acceleration.
The average acceleration of an object over a time interval is the change in its velocity divided by the time interval. It is a vector. SI unit: metre per second squared (m s⁻²).
The direction of acceleration tells a story:
- If velocity is increasing, acceleration acts in the direction of motion (positive).
- If velocity is decreasing, acceleration acts opposite to motion (negative — often called retardation or deceleration).
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Metre per second squared (m s⁻²).Show Answer
a = (v − u) ÷ t = (20 − 5) ÷ 3 = 15 ÷ 3 = 5 m s⁻², acting in the direction of motion.Show Answer
Distance = 200 m, time = 20 s → average speed = 10 m s⁻¹. Displacement = 0 (returns to start) → average velocity = 0 m s⁻¹. Direction reversal makes them differ completely.Show Answer
Zero. Acceleration measures the change in velocity. Since neither the magnitude nor the direction of velocity changes, the change is zero → a = 0, regardless of how fast the bus goes.Words and numbers describe motion, but a graph lets us see it at a glance. A motion graph shows how one quantity (position, velocity or acceleration) changes against another (usually time). From a single graph you can compare two objects, read off values, calculate other quantities, and instantly tell whether motion is uniform or accelerated. In the exam, graph questions are guaranteed marks once you learn to read the shapes.
To plot a graph we take graph paper, draw the horizontal X-axis and vertical Y-axis meeting at the origin O, choose a suitable scale for each axis, mark the points from a data table, and join them. Here is the data for a vehicle moving on a straight road:
| Time | 0 s | 1 s | 2 s | 3 s | 4 s | 5 s | 6 s |
|---|---|---|---|---|---|---|---|
| Position | 0 m | 20 m | 40 m | 60 m | 80 m | 100 m | 120 m |
Plotting time on the X-axis and position on the Y-axis, and joining the points, gives a perfectly straight line:
The line is straight and rising. Equal distances are covered in equal times → this is constant velocity. The steeper such a line, the faster the object.
Not all motion gives a straight line. When a vehicle starts from rest and keeps speeding up, its position data curves:
| Time | 0 s | 2 s | 4 s | 6 s | 8 s | 10 s | 12 s |
|---|---|---|---|---|---|---|---|
| Position | 0 m | 1 m | 4 m | 9 m | 16 m | 25 m | 36 m |
This line curves upward and steepens. More distance is covered in each successive equal time interval → velocity is increasing. Compare it directly with the straight line above to feel the difference.
The shape of a position-time graph reveals the nature of motion at a glance:
- Straight sloping line → constant velocity (uniform motion).
- Curved line → changing velocity (accelerated motion).
- Flat horizontal line → the object is at rest (position not changing).
But a position-time graph gives more than shape — it gives the velocity, through its slope. The slope of a line is its steepness: how much the Y-quantity changes for a given change in the X-quantity.
Same starting point, two slopes. The pink line B rises faster than blue line A — more displacement in the same time. Steeper = faster. This one idea answers most position-time graph questions.
A velocity-time graph shows how velocity changes with time — and it is even more powerful than the position-time graph, because it gives us two quantities: acceleration (from its slope) and displacement (from the area beneath it).
Three basic shapes cover most cases:
Memorise these three: flat = steady speed (a = 0) · up-slope = speeding up (+a) · down-slope = slowing down (−a). On a velocity-time graph, the slope IS the acceleration.
The area between the velocity-time line and the time axis equals the displacement in that interval. For a flat line it is a rectangle; for a sloping line, split it into a rectangle plus a triangle.
A velocity-time graph gives you two answers at once: slope → acceleration, area under the line → displacement. Rectangle for steady speed; rectangle + triangle when the line slopes.
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Constant velocity, which means zero acceleration.Show Answer
Acceleration = slope of the line. Displacement = area enclosed between the line and the time axis.Show Answer
The graph is a horizontal line at 20 m s⁻¹. Displacement = area of the rectangle = velocity × time = 20 × 6 = 120 m.So far we have described motion with words and graphs. Now we build the three equations that let us calculate motion exactly. These are called the kinematic equations, and they connect five quantities: initial velocity u, final velocity v, acceleration a, time t, and displacement s. They work only when acceleration is constant — which is the case throughout this chapter. Learn to derive all three; derivations are often asked directly for 3–5 marks.
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v = u + at · s = ut + ½at² · v² = u² + 2as.Show Answer
u = 0, a = 2, t = 5. v = u + at = 0 + 2×5 = 10 m s⁻¹. s = ut + ½at² = 0 + ½×2×25 = 25 m.Show Answer
Use v² = u² + 2as: 0 = 20² + 2a(100) → 0 = 400 + 200a → a = −2 m s⁻². Then v = u + at: 0 = 20 + (−2)t → t = 10 s.Everything so far has been motion in a straight line — one dimension. But a kicked football, an overtaking car, or a satellite orbiting Earth moves in a plane — this is motion in two dimensions. We will study one important two-dimensional case: circular motion.
When an object moves along a circular path, its motion is called circular motion. Think of a child on a merry-go-round moving from A to B to C. The distance travelled is the curved path ABC, but the displacement is only the straight line AC — once again, the two differ.
In one complete revolution, the object travels the whole boundary of the circle — its circumference, 2πR (R = radius). But its displacement is zero, because it ends exactly where it began. If one revolution takes time T, the average speed is:
The average velocity over one full revolution is zero (displacement is zero). When the object moves round the circle at constant speed, we call it uniform circular motion.
When an object moves in a circular path with constant (uniform) speed, its motion is called uniform circular motion.
Here is a beautiful idea. Imagine an athlete running on a square track — she changes direction 4 times per lap. On a hexagonal track, 6 times. As we keep increasing the number of sides, the turns become more frequent, and finally the track becomes a circle — where the direction of velocity changes continuously, at every single instant.
⚠️ MOST MISUNDERSTOOD POINT — read twiceTwo key facts follow:
- Acceleration is non-zero whenever velocity changes — and velocity changes if either its magnitude or its direction (or both) change.
- In uniform circular motion, the speed (magnitude) is constant, so it is only the direction that changes — and that alone is enough to make the motion accelerated.
In the real world, perfect uniform circular motion (constant speed on a perfect circle) is rare, so it is an idealised model. Still, it is extremely useful — it is the foundation for understanding planets orbiting the Sun and vehicles taking circular turns.
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Yes. The speedometer shows only speed (magnitude). If she rides along a curve or turns, the direction of velocity changes, so she is accelerating even at constant speed.Show Answer
Because velocity is a vector with both magnitude and direction. In uniform circular motion the direction of velocity changes continuously, so velocity changes even though its magnitude (speed) stays constant. A changing velocity means the motion is accelerated.Show Answer
Average speed = 2πR ÷ T = 2 × (22/7) × 7 ÷ 22 = 44 ÷ 22 = 2 m s⁻¹. Average velocity = 0 (displacement is zero after one full revolution).- Position = distance and direction of an object from a reference point at an instant.
- An object is in motion if its position changes with time; at rest if it does not.
- Displacement = net change in position between two instants (a vector).
- Average speed = total distance ÷ time interval (a scalar).
- Average velocity = displacement ÷ time interval (a vector).
- Average acceleration = change in velocity ÷ time interval (a vector).
- Three kinematic equations (constant acceleration): v = u + at · s = ut + ½at² · v² = u² + 2as.
- Motion in a circular path at constant speed = uniform circular motion — it is accelerated because direction changes.
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He covers the 250 m stretch four times: 250 × 4 = 1000 m. He ends at home where he started → displacement = 0 m.Show Answer
Ground→4th = 4 × 3 = 12 m up. 4th→2nd = 2 × 3 = 6 m down. (i) Total distance = 12 + 6 = 18 m. (ii) Final position = 2nd floor = 6 m above the start (upward displacement).Show Answer
Yes. The speedometer shows only speed (magnitude). If she turns or rides along a curve, the direction of velocity changes, so the scooter accelerates even at constant speed.Show Answer
a = (24 − 0) ÷ 6 = 4 m s⁻². s = ut + ½at² = 0 + ½ × 4 × 6² = ½ × 4 × 36 = 72 m.Show Answer
v² = u² + 2as: 0 = 28² + 2a(98) → 0 = 784 + 196a → a = −4 m s⁻². Then v = u + at: 0 = 28 + (−4)t → t = 7 s.Show Answer
Velocity = slope of the position-time line. The two lines have different slopes, so their velocities are never equal — not even where the lines cross (that only means they are at the same position at that moment, not the same velocity). Answer: No.Show Answer
Both have the same initial and final positions, so over 10 s their displacement is equal → same average velocity, and (one direction) same average speed. Correct: (i) average velocities equal and (ii) average speeds equal. Options (iii) and (iv) are wrong.Show Answer
u = 15 m s⁻¹, v = 10 m s⁻¹, t = 36 s. Distance = average velocity × time = ½(u + v) × t = ½(15 + 10) × 36 = ½ × 25 × 36 = 450 m.Show Answer
Phase 1 = ½(0 + 20) × 5 = 50 m. Phase 2 = 20 × 10 = 200 m. Phase 3 = ½(20 + 0) × 6 = 60 m. Total = 50 + 200 + 60 = 310 m.Show Answer
u = 10 m s⁻¹. Reaction distance = 10 × 0.5 = 5 m. Braking distance: 0 = 10² + 2(−2.5)s → s = 20 m. Total = 5 + 20 = 25 m. Since 25 m < 30 m, yes, the bus stops before the obstacle (with 5 m to spare).Show Answer
Rest and motion are relative — they depend on the reference point. Relative to Earth’s surface, the object is at rest. Relative to the Sun, it moves with the Earth. Both statements are correct for their chosen reference point.Show Answer
Displacement = area = triangle(0–20) + rectangle(20–100) + trapezium(100–120) = ½×20×3 + 80×3 + ½(3+2)×20 = 30 + 240 + 50 = 320 m. Average acceleration over 120 s = (2 − 0) ÷ 120 ≈ 0.017 m s⁻². (Shade the 20–100 s rectangle for constant velocity, and the 100–120 s part for decreasing velocity.)Show Answer
Distance = area under the graph ≈ average velocity × time ≈ 7 km h⁻¹ × 6 h ≈ 42 km (a marathon distance). Accept ~40–45 km depending on how the graph is read.Show Answer
Phase 1 = 6 × 120 = 720 m. Phase 2 = ut + ½at² = 6×6 + ½×1×36 = 36 + 18 = 54 m. Total = 720 + 54 = 774 m.Show Answer
aA = 5 ÷ 5 = 1 m s⁻²; aB = 3 ÷ 10 = 0.3 m s⁻². Displacement A (5 s) = ½ × 5 × 5 = 12.5 m. Displacement B (10 s) = ½ × 10 × 3 = 15 m.Show Answer
Time = 1.5 h = 5400 s = 1.5 revolutions. R = 0.07 m; circumference = 2πR ≈ 0.44 m. (i) Distance = 1.5 × 0.44 ≈ 0.66 m. (ii) After 1.5 revolutions the hand is opposite its start → displacement = diameter = 2R = 0.14 m. (iii) Speed = 0.66 ÷ 5400 ≈ 1.2 × 10⁻⁴ m s⁻¹. (iv) Velocity = 0.14 ÷ 5400 ≈ 2.6 × 10⁻⁵ m s⁻¹, directed from start to end position.Show Answer
Write numbers 1–12 on the outer ring (7 cm) and letters A–F on the inner ring (4 cm), then spin. The outer marks travel a bigger circle, so they move at a higher speed (v = 2πR ÷ T, larger R → larger v). Faster marks blur and fade first, while the slower inner letters stay readable. The speeds are different — outer faster than inner.Show Answer
On an outstretched palm the app shows tiny fluctuating readings (small involuntary hand movements = tiny accelerations); on the floor the readings are nearly zero (no motion). This shows real objects are rarely perfectly at rest. Such tiny movements are studied in medical research, e.g. movement disorders.Show Answer
s = ½(u + v)t: the velocity-time graph is a trapezium with parallel sides u and v and width t. Area of trapezium = ½(sum of parallel sides) × width = ½(u + v)t, and area = displacement, so s = ½(u + v)t.s = vt − ½at²: start from s = ut + ½at² and substitute u = v − at (from v = u + at): s = (v − at)t + ½at² = vt − at² + ½at² = vt − ½at².
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Scale: plotting the same data with different axis scales makes the line look steep or flat — so always read the axis labels before judging a graph. Mechanic activity: braking distance increases with wet roads, worn tyres, higher vehicle mass, night/fog driving, severe weather, and longer reaction time — good material for a school road-safety poster and skit.📖 All Definitions — write these word-for-word
The distance and direction of an object with respect to a reference point at a given instant describes its position.
The net change in the position of an object between two given instants of time (a vector). SI unit: metre.
Total distance travelled divided by the time interval (a scalar). SI unit: m s⁻¹.
Displacement divided by the time interval (a vector). SI unit: m s⁻¹.
Change in velocity divided by the time interval (a vector). SI unit: m s⁻².
Equal distances in equal times = uniform; unequal distances in equal times = non-uniform.
Motion in a circular path with constant speed; it is accelerated because the direction of velocity changes.
Scalar = magnitude only (distance, speed). Vector = magnitude + direction (displacement, velocity, acceleration).
⚡ ALL FORMULAS — ONE-GLANCE REVISION
- Distance vs displacement: never write them as equal when the object turns back. Displacement can be less; distance never decreases.
- Sign of acceleration: a negative sign is not “wrong” — it shows the direction (slowing down / opposite to motion). Always keep the minus sign.
- Unit conversion: forgetting to convert km h⁻¹ to m s⁻¹ (÷ 3.6) before using formulas is the #1 numerical error.
- Graph confusion: a flat line on a position-time graph = at rest; a flat line on a velocity-time graph = constant speed. Different meanings.
- Slope vs area: on a velocity-time graph, slope = acceleration, area = displacement. Don’t swap them.
- Circular motion: constant speed still means accelerating, because direction changes. This is the most-tested trap.
- Distance = whole path (scalar); Displacement = start-to-end straight line with direction (vector).
- Speed uses distance; Velocity uses displacement. Both in m s⁻¹.
- Acceleration = how fast velocity changes = (v − u) ÷ t, in m s⁻².
- Pick the equation by what’s missing: no s → v=u+at · no v → s=ut+½at² · no t → v²=u²+2as.
- Position-time: slope = velocity; straight = constant velocity; curve = accelerating; flat = at rest.
- Velocity-time: slope = acceleration; area = displacement.
- Uniform circular motion = constant speed but accelerating (direction changes); average velocity over one round = 0.
- Always convert km h⁻¹ → m s⁻¹ (÷ 3.6) and keep the sign of acceleration.