Describing Motion Around Us — Detailed Notes
Everything in nature is in motion. To study complex motions we use idealised forms: linear (straight-line), circular and oscillatory. This chapter covers motion in a straight line and uniform circular motion, described in words, numbers, equations and graphs.
4.1 Position, distance & displacement
Position is given by the distance and direction of an object from a fixed reference point (origin). An object is in motion if its position changes with time, otherwise it is at rest. On a line, direction is shown by + and − signs.
Distance travelled = total path length covered (only magnitude, SI unit m). Displacement = net change in position between two instants — it has both magnitude and direction. Magnitude of displacement ≤ total distance; they are equal only when the object moves in one direction without turning back. Quantities with only magnitude are scalars; those needing magnitude + direction are vectors.
4.1.3 Average speed & average velocity
Average speed = total distance ÷ time interval (scalar). Average velocity vav = displacement ÷ time interval = s ÷ t (vector). SI unit of both: m s⁻¹ (also km h⁻¹; 1 km h⁻¹ = 5/18 m s⁻¹). Uniform motion = equal distances in equal time intervals (constant speed); non-uniform = unequal distances in equal times.
4.1.4 Average acceleration
Average acceleration a = change in velocity ÷ time = (v − u) ÷ (t₂ − t₁). SI unit: m s⁻². If speed increases, a is in the direction of velocity (+); if speed decreases, a is opposite (−, called retardation). Acceleration can arise from a change in magnitude or direction of velocity, or both. An object can move fast yet have zero acceleration (constant velocity). Freely falling bodies have constant acceleration g = 9.8 m s⁻² downward.
4.2 Graphs of motion
Position–time graph: a straight line → constant velocity; the slope = velocity. A line parallel to the time axis → object at rest. A curve → changing velocity (acceleration). Steeper slope → greater velocity.
Velocity–time graph: the slope = acceleration; the area under the graph (with the time axis) = displacement. A horizontal line → constant velocity (zero acceleration); a sloping straight line → constant acceleration.
4.3 Kinematic equations (constant acceleration)
With u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement:
(1) v = u + at
(2) s = ut + ½at²
(3) v² = u² + 2as
These apply only when acceleration is constant. Signs of u, v, a, s indicate direction.
4.4 Motion in a plane & uniform circular motion
Motion in a plane is two-dimensional (e.g. a kicked ball, a satellite). In circular motion, in one revolution the distance = circumference = 2πR, but displacement = 0 (returns to start). Average speed for one revolution vav = 2πR ÷ T. Uniform circular motion = circular path at constant speed. Even though the speed is constant, the direction of velocity changes continuously (velocity is along the tangent), so the motion is accelerated.
Key Terms
reference point, position, distance, displacement, scalar, vector, average speed, average velocity, average acceleration, g, position–time graph, velocity–time graph, slope, kinematic equations, uniform circular motion, tangent.
Revise, Reflect, Refine — All NCERT Exercise Answers
1Father goes 250 m to a shop, returns home for a bag, goes to the shop again, buys and comes home. Total distance? Displacement from home?
Path: home→shop (250 m) → home (250 m) → shop (250 m) → home (250 m). Total distance = 250 × 4 = 1000 m. Since he starts and ends at home, the net change in position is zero, so displacement = 0 m.
2A student runs from the ground floor to the 4th floor, then down to the 2nd floor. Each floor is 3 m. (i) total vertical distance (ii) displacement.
Ground→4th = 4 × 3 = 12 m up; 4th→2nd = 2 × 3 = 6 m down. (i) Total distance = 12 + 6 = 18 m. (ii) Displacement = position of 2nd floor from ground = 2 × 3 = 6 m (upward).
3A girl’s scooter speedometer reads constant. Can the scooter be accelerating? How?
Yes. The speedometer shows only the magnitude of velocity (speed). If the scooter is turning (changing direction), the direction of its velocity changes even though the speed is constant — and any change in velocity means acceleration. So on a curved path at constant speed, the scooter is accelerating.
4A car starts from rest and reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in 6 s.
u = 0, v = 24 m s⁻¹, t = 6 s. a = (v − u)/t = 24/6 = 4 m s⁻². Distance s = ut + ½at² = 0 + ½ × 4 × 6² = ½ × 4 × 36 = 72 m (or s = average velocity × t = 12 × 6 = 72 m).
5A motorbike with u = 28 m s⁻¹ and constant acceleration stops after 98 m. Find the acceleration and time to stop.
v = 0, u = 28 m s⁻¹, s = 98 m. Using v² = u² + 2as: 0 = 28² + 2a(98) → 0 = 784 + 196a → a = −784/196 = −4 m s⁻² (retardation). Time: v = u + at → 0 = 28 + (−4)t → t = 7 s.
6Position–time graphs of A and B on parallel tracks, same direction (Fig. 4.27). Do A and B ever have equal velocity?
On a position–time graph, velocity = slope. The two objects have equal velocity at the instant where the slopes of their graphs are equal — i.e. where the tangents to the two curves are parallel. At that instant (typically near where one is about to overtake the other) their velocities match, even though their positions differ. So yes, they can have equal velocity at that moment.
7Fig. 4.28: A and B have the SAME initial and final positions over 0–10 s. Choose the correct option(s).
(i) is correct — since both have the same displacement in the same time, their average velocities are equal. The object that follows the longer path (greater total distance) has the higher average speed; here A traces the longer path, so (iv) is also correct (A’s average speed is greater than B’s). Options (ii) and (iii) are wrong because the two do not cover equal distances.
8A truck at 54 km h⁻¹ slows to 36 km h⁻¹ in 36 s (constant acceleration). Find the distance travelled.
u = 54 km h⁻¹ = 15 m s⁻¹, v = 36 km h⁻¹ = 10 m s⁻¹, t = 36 s. Distance = average velocity × t = ((15 + 10)/2) × 36 = 12.5 × 36 = 450 m.
9A car: rest → 20 m s⁻¹ in 5 s, then 20 m s⁻¹ for 10 s, then brakes to stop in 6 s. Find total distance.
Phase 1 (accelerating): s₁ = ½(u+v)t = ½(0+20)(5) = 50 m. Phase 2 (constant): s₂ = 20 × 10 = 200 m. Phase 3 (braking): s₃ = ½(20+0)(6) = 60 m. Total distance = 50 + 200 + 60 = 310 m.
10A bus at 36 km h⁻¹ sees an obstacle 30 m ahead; reaction time 0.5 s, then brakes at 2.5 m s⁻². Will it stop in time?
u = 36 km h⁻¹ = 10 m s⁻¹. Reaction distance = 10 × 0.5 = 5 m. Braking distance: v² = u² + 2as → 0 = 10² + 2(−2.5)s → s = 100/5 = 20 m. Total stopping distance = 5 + 20 = 25 m, which is less than 30 m — yes, the bus stops about 5 m before the obstacle.
11“The Earth moves around the Sun.” Can an object kept on the Earth be considered at rest?
Rest and motion are relative to the chosen reference point. An object on the ground is at rest relative to the Earth (its position w.r.t. the Earth does not change), but it is in motion relative to the Sun (it moves with the Earth around the Sun). So the same object can be at rest or in motion depending on the reference point chosen.
12Velocity–time graph of a cyclist, 0–120 s (Fig. 4.30). Shade displacement for (i) constant-velocity and (ii) decreasing-velocity parts; find total displacement and average acceleration over 120 s.
On a v–t graph, displacement = area under the line. (i) Shade the rectangle under the flat (constant-velocity) part; (ii) shade the triangle under the sloping-down (decreasing-velocity) part. Total displacement = sum of all areas (rectangle + triangles) read off the graph. Average acceleration over 120 s = (final velocity − initial velocity) ÷ 120 s. (Read the exact velocity values from the graph axes and substitute; method as shown.)
13A runner’s velocity–time graph is given (Fig. 4.31). Estimate the distance she ran.
Distance = area between the velocity–time line and the time axis. Split the area into rectangles and triangles, calculate each (converting km h⁻¹ and hours consistently, or to m and s), and add them to get the total distance run. (Substitute the values read from the graph.)
14A car moves at constant 6 m s⁻¹ for 2 min, then accelerates at 1 m s⁻² for 6 s. Find displacement in 2 min 6 s.
Phase 1 (constant): s₁ = 6 × 120 = 720 m. Phase 2 (accelerating): s₂ = ut + ½at² = 6 × 6 + ½ × 1 × 6² = 36 + 18 = 54 m. Total displacement = 720 + 54 = 774 m. (On the v–t graph this is the rectangle + the trapezium area.)
15Car A reaches 5 m s⁻¹ in 5 s; Car B reaches 3 m s⁻¹ in 10 s (both from rest, constant acceleration). Plot v–t graphs and find displacements.
Accelerations: aA = 5/5 = 1 m s⁻²; aB = 3/10 = 0.3 m s⁻². Both v–t graphs are straight lines from the origin (A steeper). Displacement = area of the triangle: Car A (in 5 s) = ½ × 5 × 5 = 12.5 m; Car B (in 10 s) = ½ × 10 × 3 = 15 m.
16The minute hand (length 7 cm) from 6 PM to 7:30 PM. Find its (i) distance (ii) displacement (iii) speed (iv) velocity.
6:00 → 7:30 = 90 min = 1.5 revolutions; r = 7 cm, so circumference = 2πr = 2 × (22/7) × 7 = 44 cm. (i) Distance = 1.5 × 44 = 66 cm. (ii) Displacement: after 1.5 rev the tip moves from the ‘12’ to the ‘6’ position = diameter = 2r = 14 cm. (iii) Speed = 66 cm ÷ 90 min ≈ 0.73 cm min⁻¹ (≈ 0.0122 cm s⁻¹). (iv) Velocity = 14 cm ÷ 90 min ≈ 0.16 cm min⁻¹ (≈ 0.0026 cm s⁻¹), directed from the 12 toward the 6.