Describing Motion Around Us
Learn how to describe, measure and analyse motion using quantities like displacement, velocity, acceleration and graphs.
Everything around us is in motion — from planets in space down to tiny atoms. Butterflies flit, snakes slither, horses gallop, clouds gather, and ocean tides rise and fall. To make sense of all this variety, scientists study motion in its simplest forms first: linear (straight-line), circular, and oscillatory motion.
In this chapter we focus on motion in a straight line and uniform circular motion. You will move beyond distance, time and speed to understand displacement, average velocity and average acceleration, and learn to represent motion with equations and graphs.
How much distance should you keep from the truck ahead to avoid a collision if it suddenly brakes? Does that distance depend on how fast you are moving?
4.1 Motion in a Straight Line
When an object moves along a straight path, we call it linear motion. It is the simplest type of motion. Examples include swimmers in a race lane, a falling ball, a car on a straight highway, and a train on a straight track.
To describe motion, we first need to track the position of an object at different moments of time.
4.1.1 Describing Position
Position is described by stating how far and in which direction an object is from a fixed reference point (also called the origin).
Motion: If the position of an object changes with time relative to a reference point, the object is in motion.
Rest: If position does not change with time relative to a reference point, the object is at rest.
For straight-line motion, we use + (positive) for one direction (usually right/upward) and − (negative) for the opposite direction.
4.1.2 Distance Travelled and Displacement
Distance Travelled
The total length of the path covered by an object. It only has a numerical value (no direction). It can never be negative.
Displacement
The net change in position between two instants of time. It has both a numerical value (magnitude) and a direction. It can be zero or negative.
The magnitude of displacement is always less than or equal to the total distance travelled. They are equal only when the object moves in one direction without turning back. The SI unit of both is the metre (m).
4.1.3 Average Speed and Average Velocity
Average Speed
Average speed tells us how fast or slow an object is moving overall. It has no direction.
Uniform Motion
Equal distances in equal time intervals → constant speed.
Non-Uniform Motion
Unequal distances in equal time intervals → changing speed.
Average Velocity
Average velocity tells us how fast the position is changing and in which direction. It requires both magnitude and direction.
Average velocity is the rate of change of position. Its direction is the same as the direction of displacement (shown by + or − sign). The SI unit of both average speed and average velocity is m s⁻¹ (also written m/s) or km h⁻¹.
4.1.4 Average Acceleration
When a vehicle starts, stops, or changes speed, its velocity changes. The quantity that describes how quickly velocity changes is acceleration.
The average acceleration of an object over a time interval is the change in velocity divided by the time interval.
where u = initial velocity, v = final velocity, and the denominator is the time interval. The SI unit of acceleration is m s⁻².
An object can be moving very fast and still have zero acceleration if its velocity is not changing. Acceleration depends on how quickly velocity changes, not on how fast the object moves. A bus at constant velocity on a highway has zero acceleration.
4.2 Graphical Representation of Motion
Graphs give us a visual picture of how position, velocity, or acceleration change with time. They help us compare motions, read off values, and identify whether motion is uniform or non-uniform.
4.2.2 Position-Time Graphs
Slope of position-time graph = velocity. A steeper slope means higher velocity. A horizontal line (slope = 0) means the object is stationary.
4.2.3 Velocity-Time Graphs
- Slope of v-t graph = acceleration
- Area under v-t graph (between line and time axis) = displacement
4.3 Kinematic Equations for Straight-Line Motion with Constant Acceleration
When acceleration is constant, we can link the five quantities — displacement (s), time (t), initial velocity (u), final velocity (v), and acceleration (a) — using three equations.
Deriving the Equations
Equation 1: Velocity-Time Relation
Starting from the definition of acceleration:
a = (v − u) / t
⟹ at = v − u
Equation 2: Position-Time Relation
Displacement equals the area under the v-t graph (trapezium with parallel sides u and v, width t):
s = area of trapezium = ut + ½at²
Equation 3: Position-Velocity Relation
Eliminate t by substituting t = (v − u)/a into Equation 2:
- These equations are valid only when acceleration is constant.
- For one-direction motion: distance = |s|, speed = |v|.
- The + or − sign of u, v, a, s tells you the direction of each quantity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| u | Initial velocity | m s⁻¹ |
| v | Final velocity | m s⁻¹ |
| a | Acceleration | m s⁻² |
| t | Time interval | s |
| s | Displacement | m |
4.4 Motion in a Plane
Motion along a 2D path — such as a car turning, a football being kicked, or a satellite orbiting — is called motion in two dimensions (a plane).
4.4.1 Uniform Circular Motion
When an object moves along a circular path at constant (uniform) speed, the motion is called uniform circular motion.
Average Speed in Circular Motion
For one complete revolution of radius R completed in time period T:
The average velocity for one complete revolution is zero because the displacement is zero (the object returns to its starting point).
Speed in UCM
Constant — same magnitude at every point on the circle. Equal to 2πR/T.
Velocity in UCM
Changing direction continuously — tangent to the circle at every point. Never constant.
In uniform circular motion, the object is always accelerating — even though speed is constant — because the direction of velocity keeps changing. This is why we say acceleration can arise from a change in direction alone, not just a change in speed.
Practice Questions & Solutions
My father went to a shop 250 m away on a straight road. On reaching, he found he forgot his cloth bag. He came home, took it, went to the shop again, bought provisions and came back home. What is the total distance travelled and his displacement from home?
Trips made:
- Home → Shop: 250 m
- Shop → Home: 250 m
- Home → Shop: 250 m
- Shop → Home: 250 m
Total distance = 250 × 4 = 1000 m
Displacement = 0 m (he starts and ends at home, so no net change in position)
A student runs from the ground floor to the 4th floor to collect a book, then comes down to the 2nd floor classroom. Height of each floor = 3 m. Find (i) total vertical distance travelled, (ii) displacement from starting point.
(i) Total vertical distance:
- Ground (0 m) to 4th floor = 4 × 3 = 12 m (up)
- 4th floor to 2nd floor = 2 × 3 = 6 m (down)
- Total distance = 12 + 6 = 18 m
(ii) Displacement:
Starting position = 0 m (ground), Final position = 2nd floor = 2 × 3 = 6 m
Displacement = 6 − 0 = 6 m upward
A girl rides her scooter and the speedometer reading stays constant. Is it possible for the scooter to be accelerating? If so, how?
Yes, it is possible. The speedometer shows speed (magnitude of velocity), not velocity itself. If the girl takes a turn or moves in a curved path, the direction of her velocity changes even though the speed is constant. Since acceleration is defined as any change in velocity (including change in direction), the scooter is accelerating even with a constant speedometer reading.
This is an example of uniform circular motion — speed is constant but velocity (and hence acceleration) is not zero.
A car starts from rest and reaches a velocity of 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in 6 s.
Given: u = 0, v = 24 m s⁻¹, t = 6 s
Average acceleration:
a = (v − u) / t = (24 − 0) / 6 = 4 m s⁻²
Distance travelled:
s = ut + ½at² = 0 × 6 + ½ × 4 × 36 = 0 + 72 = 72 m
A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration and the time taken to stop.
Given: u = 28 m s⁻¹, v = 0 (stops), s = 98 m
Finding acceleration using v² = u² + 2as:
0 = (28)² + 2 × a × 98
0 = 784 + 196a
a = −784 / 196 = −4 m s⁻² (negative = deceleration)
Finding time using v = u + at:
0 = 28 + (−4) × t
t = 28 / 4 = 7 s
Fig. 4.27 shows a position-time graph of two objects A and B moving along parallel tracks in the same direction. Do A and B ever have equal velocity?
From the graph, both A and B have straight-line position-time graphs (meaning constant velocity each). The slope of A's line appears steeper than B's, meaning A has higher velocity than B. Since both slopes (velocities) are constant and different, A and B never have equal velocity during the motion shown.
Equal velocity would mean equal slopes — i.e., lines parallel to each other. As these lines are not parallel, their velocities are never equal.
Fig. 4.28 shows position vs time for objects A and B (0 to 10 s). Choose the correct option(s):
(i) Average velocity of both over 10 s is equal since they have the same initial and final positions.
(ii) Average speeds are equal since both cover equal distance in equal time.
(iii) Average speed of A is lower than B as it covers shorter distance.
(iv) Average speed of A is greater than B since B's speed is lower in some segments.
Correct options: (i) and (iv)
(i) ✓ — Average velocity = displacement / time. If both A and B start and end at the same position, their displacement is the same, so average velocity over 10 s is equal.
(ii) ✗ — A has a curved path (it moves faster in parts and slower in others) while B has a straight path. A actually covers more total path length to reach the same final position if it overshoots and comes back — their distances are NOT necessarily equal.
(iii) ✗ — This would be true only if A had a simpler, shorter path. From the graph, A (curved) may cover more distance than B.
(iv) ✓ — From the graph, A's curve is steeper at some parts, indicating higher speed in those segments; B's straight line has constant (lower average speed in places). So A's average speed can be greater.
A truck driver at 54 km h⁻¹ slows to 36 km h⁻¹ in 36 s. Find the distance travelled during this time (constant acceleration).
Convert to m s⁻¹: u = 54 km h⁻¹ = 15 m s⁻¹, v = 36 km h⁻¹ = 10 m s⁻¹, t = 36 s
Using s = ½(u + v) × t:
s = ½ × (15 + 10) × 36 = ½ × 25 × 36 = 450 m
A car starts from rest, accelerates uniformly to 20 m s⁻¹ in 5 s, travels at 20 m s⁻¹ for 10 s, then brakes uniformly to stop in 6 s. Find the total distance.
Phase 1 (acceleration, 0 to 5 s):
u = 0, v = 20 m s⁻¹, t = 5 s
s₁ = ½(u + v)t = ½(0 + 20) × 5 = 50 m
Phase 2 (constant velocity, 5 to 15 s):
s₂ = 20 × 10 = 200 m
Phase 3 (braking, 15 to 21 s):
u = 20 m s⁻¹, v = 0, t = 6 s
s₃ = ½(20 + 0) × 6 = 60 m
Total distance = 50 + 200 + 60 = 310 m
A bus at 36 km h⁻¹ sees an obstacle 30 m ahead. Driver reaction time = 0.5 s. After braking, acceleration = −2.5 m s⁻². Will the bus stop in time?
Convert: 36 km h⁻¹ = 10 m s⁻¹
Distance during reaction time (0.5 s at constant speed):
s₁ = 10 × 0.5 = 5 m
Distance while braking (u = 10 m s⁻¹, v = 0, a = −2.5 m s⁻²):
v² = u² + 2as → 0 = 100 + 2 × (−2.5) × s₂
s₂ = 100 / 5 = 20 m
Total distance = 5 + 20 = 25 m
Since 25 m < 30 m, yes, the bus stops safely before the obstacle (5 m to spare).
"The Earth moves around the Sun." In this context, can an object on Earth be considered at rest?
This depends on the choice of reference point. Motion and rest are always relative.
- If we take the Sun as the reference point: the Earth (and everything on it) is in motion — moving around the Sun.
- If we take a point on the Earth's surface as the reference point: an object sitting on a table does not change its position relative to the table or floor. So it is at rest relative to Earth.
Both descriptions are correct — rest and motion are relative and depend on the reference point chosen.
A velocity-time graph for a cyclist from 0 s to 120 s is given (velocity increases from 0 to 3 m s⁻¹ in 40 s, stays at 3 m s⁻¹ from 40 to 80 s, then decreases to 0 in 120 s). Find total displacement and average acceleration over 120 s.
Displacement = Total area under the v-t graph:
- Phase 1 (0–40 s, triangle): Area = ½ × 40 × 3 = 60 m
- Phase 2 (40–80 s, rectangle): Area = 40 × 3 = 120 m
- Phase 3 (80–120 s, triangle): Area = ½ × 40 × 3 = 60 m
Total displacement = 60 + 120 + 60 = 240 m
Average acceleration over 120 s:
a = (v − u) / t = (0 − 0) / 120 = 0 m s⁻²
(The cyclist starts and ends at the same speed, so the average acceleration over the entire interval is zero.)
A girl running for a marathon has a velocity-time graph (Fig. 4.31) where velocity increases from 0 to 7.5 km h⁻¹ in 2 h, stays constant till 4 h, then decreases to 5 km h⁻¹ by 6 h. Estimate the running distance.
Convert velocities: 7.5 km h⁻¹ stays in km/h; time in hours → distance in km.
- Phase 1 (0–2 h, triangle): ½ × 2 × 7.5 = 7.5 km
- Phase 2 (2–4 h, rectangle at 7.5 km/h): 2 × 7.5 = 15 km
- Phase 3 (4–6 h, trapezium, v goes from 7.5 to 5 km/h): ½ × (7.5 + 5) × 2 = 12.5 km
Total running distance ≈ 7.5 + 15 + 12.5 = 35 km
A car enters a highway at 6 m s⁻¹ and travels at constant velocity for 2 min, then accelerates at 1 m s⁻² for 6 s. Find total displacement in 2 min 6 s.
Phase 1 (constant velocity, 120 s):
s₁ = 6 × 120 = 720 m
Phase 2 (acceleration for 6 s, u = 6 m s⁻¹, a = 1 m s⁻²):
s₂ = ut + ½at² = 6 × 6 + ½ × 1 × 36 = 36 + 18 = 54 m
Total displacement = 720 + 54 = 774 m
Car A reaches 5 m s⁻¹ in 5 s from rest; Car B reaches 3 m s⁻¹ in 10 s from rest. Plot their v-t graphs and find displacement for each in their respective time intervals.
Car A: u = 0, v = 5 m s⁻¹, t = 5 s
Acceleration aₐ = (5−0)/5 = 1 m s⁻²
Displacement sₐ = ½ × (0 + 5) × 5 = 12.5 m
Car B: u = 0, v = 3 m s⁻¹, t = 10 s
Acceleration a_b = (3−0)/10 = 0.3 m s⁻²
Displacement s_b = ½ × (0 + 3) × 10 = 15 m
Rohan studies from 6 PM to 7:30 PM (1.5 hours = 90 min). The minute hand of a clock has length 7 cm. Find: (i) distance travelled, (ii) displacement, (iii) speed, (iv) velocity.
In 90 minutes, the minute hand completes exactly 1.5 revolutions (one full revolution per 60 minutes).
(i) Distance travelled:
Circumference of one full revolution = 2πr = 2 × 3.14 × 7 = 43.96 cm
For 1.5 revolutions: d = 1.5 × 43.96 = 65.94 cm ≈ 65.9 cm
(ii) Displacement:
At 6:00 PM, the minute hand is at the 12 o'clock position.
After 90 min (at 7:30 PM), it is again at the 6 o'clock position (pointing straight down).
The tip has moved from 12 to 6, which is directly across the centre.
Displacement = diameter = 2r = 2 × 7 = 14 cm, directed from 12 toward 6 (downward)
(iii) Speed:
Time = 90 min = 90 × 60 = 5400 s
Speed = distance / time = 65.94 / 5400 ≈ 0.0122 cm s⁻¹ ≈ 1.22 × 10⁻² cm s⁻¹
(iv) Velocity (average):
Magnitude = displacement / time = 14 / 5400 ≈ 2.59 × 10⁻³ cm s⁻¹, directed from the 12 position toward the 6 position (downward from centre).
Additional Kinematic Equations (The Journey Beyond)
From the two primary equations v = u + at and s = ut + ½at², two more equations can be derived:
In a v-t graph, when velocity changes from u to v over time t, the area under the graph is a trapezium. Using the formula: Area = ½ × (sum of parallel sides) × height = ½ × (u + v) × t. Since this area equals displacement, s = ½(u + v)t.