Describing Motion Around Us teaches how to describe straight-line motion with numbers, equations and graphs. It starts with position and a reference point, separates distance travelled from displacement, and defines average speed, average velocity and average acceleration. It then shows how the slope and area of position-time and velocity-time graphs give velocity, acceleration and displacement, derives the kinematic equations for constant acceleration, and ends with uniform circular motion. Its numericals on graphs and kinematic equations are the base for Class 10 physics.
Key Concepts
1. Position, Distance and Displacement
To study the many motions in nature, scientists begin with idealised simple forms. This chapter takes two of them: linear motion (motion in a straight line, the simplest kind, such as a ball falling vertically or a train on a straight track) and uniform circular motion.
To say where an object is, you first fix a reference point. The position of an object at an instant is its distance and direction from the reference point at that instant. An object is in motion if its position with respect to the reference point changes with time, and at rest if it does not. Rest and motion depend on the reference point: a book on your desk is at rest with respect to the room and moves with the Earth around the Sun.
Direction on a straight line: mark the reference point as the origin O on a number line. Positions to the right of O are generally taken as positive (+) and to the left as negative (−). You may choose the origin and the positive direction to suit the problem, but once chosen they must not change until the problem is finished.
Instant and time interval: an instant is a single clock reading, such as t = 4 s. A time interval is the duration between two instants, such as t = 4 s to t = 10 s, which is 6 s.
Distance travelled and displacement
NCERT follows an athlete on a straight track. She starts at O at t = 0 s, passes B (40 m from O) at t = 4 s, reaches A (100 m) at t = 10 s, then runs back to B, reaching it at t = 16 s.
- Total distance travelled = OA + AB = 100 m + 60 m = 160 m, the full length of the path. It needs only a number with a unit.
- Displacement = OB = 40 m in the positive direction.
Displacement is the net change in position of an object between two given instants. Its magnitude (numerical value with unit) is the distance between the two positions, and its direction points from the first position towards the second. The SI unit of both distance and displacement is the metre (m). Quantities given by a numerical value alone are scalars (distance, speed); quantities that also need a direction are vectors (displacement, velocity, acceleration).
Key rule: magnitude of displacement ≤ distance travelled. For straight-line motion the two are equal only if the object moves in one direction without turning back.
Activity 4.1 (ball thrown up): a ball is thrown vertically upward from O on the ground. It rises in a straight line to its highest point B, then falls back along the same line to O, so this is motion in a straight line. At point A, 40 cm above O on the way up, the distance travelled is 40 cm and the displacement is 40 cm upward.
- Going up to B: the ball moves in one direction, so distance travelled and magnitude of displacement stay equal, both equal to the height OB at B.
- Coming down from B: distance travelled keeps increasing, while the upward displacement decreases, so the two become unequal.
- Back at O: distance travelled = 2 × OB, and displacement = 0.
Of the options in the activity, the true one is: the magnitude of displacement is less than or equal to the total distance travelled.
So displacement can be zero, and its magnitude never exceeds the distance travelled. Fuel used by a vehicle depends on the total distance travelled: a trip to a market 10 km away and back burns fuel for 20 km though the displacement is zero.
2. Average Speed and Average Velocity
Average speed = total distance travelled / time interval (Eq. 4.1). Distance has no direction, so average speed has only a numerical value.
- Uniform motion in a straight line: equal distances in equal intervals of time, for all possible choices of intervals. The speed is constant. The vehicle of Table 4.3 covers 20 m every second (0, 20, 40 … 120 m at 0 to 6 s).
- Non-uniform motion: unequal distances in equal intervals. If the distances in successive equal intervals keep increasing, the speed is increasing.
India’s scientific contribution: speed as distance divided by time appears in the Aryabhatiya (5th century CE). The Ganitakaumudi (14th century CE) asks when two postmen 210 yojanas apart, walking towards each other at 9 and 5 yojanas per day, will meet. They close 14 yojanas a day, so they meet after 210/14 = 15 days, having walked 135 and 75 yojanas.
Average velocity tells how fast the position changes and in which direction. It is the displacement divided by the time interval in which it occurs.
Average velocity = displacement / time interval, vav = s / t (Eq. 4.2a, 4.2b)
- Its direction is the direction of displacement, shown on a straight line by a + or − sign.
- SI unit of both average speed and average velocity: m s−1 (m/s); km h−1 is also common.
- Rate of change is the change in a quantity divided by the corresponding change in time. Average velocity is the average rate of change of position.
km/h to m/s: 1 km/h = 1000 m / 3600 s = 5/18 m/s.
| km/h | 18 | 36 | 54 | 72 | 90 | 108 |
|---|---|---|---|---|---|---|
| m/s | 5 | 10 | 15 | 20 | 25 | 30 |
NCERT Example 4.2: Sarang swims to the far end of a 25 m pool and back in 50 s. Average speed = 50 m / 50 s = 1 m/s. Displacement is 0, so average velocity = 0 m/s.
For straight-line motion, average speed equals the magnitude of average velocity if the object moves in one direction. The average velocity is zero while the average speed is not whenever the object returns to its start.
Velocity at an instant: in this chapter “velocity” alone means the velocity at an instant. As the time interval around an instant is made smaller and smaller, the average velocity approaches a fixed value, the instantaneous velocity. A speedometer reads nearly its magnitude, and the direction of the tyres gives its direction.
3. Average Acceleration
The jolt you feel when a vehicle suddenly starts or stops is the feeling of a change in velocity. The average acceleration of an object over a time interval is the change in its velocity divided by the time interval. If the velocity changes from u at time t1 to v at time t2:
a = (v − u) / (t2 − t1) (Eq. 4.3c)
- SI unit: m s−2 (m/s2). Acceleration has magnitude and direction.
- If the magnitude of velocity is increasing, the acceleration is along the velocity.
- If the magnitude of velocity is decreasing, the acceleration is opposite to the velocity; with velocity taken positive, a comes out negative.
- Acceleration can come from a change in the magnitude of velocity, its direction, or both.
Fast motion with zero acceleration: acceleration measures how quickly velocity changes. A bus on a straight highway at a steady 90 km/h has zero acceleration.
NCERT Example 4.3: a bus speeds up from 36 km/h (10 m/s) to 54 km/h (15 m/s) in 10 s: a = (15 − 10)/10 = 0.5 m/s2, along the velocity. It then brakes from 15 m/s to rest in 5 s: a = (0 − 15)/5 = −3 m/s2, opposite to the velocity.
Constant acceleration: for motion in a straight line in one direction, if velocity changes by equal amounts in equal intervals of time (for all choices of intervals), the acceleration is constant. This chapter treats only constant acceleration.
NCERT Example 4.4 (falling object): an object dropped from rest (origin at the release point, downward positive) has velocities 0, 9.8, 19.6, 29.4 and 39.2 m/s at t = 0, 1, 2, 3 and 4 s. In every 1 s interval the velocity rises by 9.8 m/s, so the average acceleration is 9.8 m/s2 in each interval. This constant downward acceleration is the acceleration due to the gravitational force of the Earth, written g. Acceleration at a single instant is called instantaneous acceleration.
4. Position-Time Graphs
Graphs show how position or velocity changes with time, help compare two motions, and separate uniform from non-uniform motion. All graphs in this chapter are for straight-line motion in one direction, where distance equals the magnitude of displacement and speed equals the magnitude of velocity.
Plotting (Activity 4.3): draw the X-axis (time) and Y-axis (position) meeting at origin O; choose a scale that uses the paper well (for Table 4.3: 5 divisions = 1 s on X, 5 divisions = 20 m on Y); plot each (time, position) pair; join the points. Table 4.3 gives a straight line. A graph shows how the position changes with time with respect to the origin. It does not show the route.
Example 4.5 data: a vehicle starting from rest has positions 0, 1, 4, 9, 16, 25 and 36 m at 0, 2, 4, 6, 8, 10 and 12 s. These points lie on a curve that bends upward. (They fit s = ½ × 0.5 × t2, a constant acceleration of 0.5 m/s2.)
| Shape of position-time graph | Meaning |
|---|---|
| Straight line parallel to the time axis | At rest (NCERT Example 4.6: a vehicle resting 40 m from the origin) |
| Straight sloping line | Constant velocity, zero acceleration |
| Curve getting steeper | Velocity increasing: accelerated motion |
| Steeper of two straight lines | Higher velocity (NCERT Example 4.7: B covers more displacement than A in the same interval) |
Velocity from slope (Activity 4.4): take points A and B on the line. Draw a line from A parallel to the X-axis and from B parallel to the Y-axis, meeting at C. BC is the change in position (s2 − s1) and CA the change in time (t2 − t1).
v = (s2 − s1) / (t2 − t1) = BC / CA
With A at (2 s, 40 m) and B at (4 s, 80 m): v = 40/2 = 20 m/s. BC/CA is the slope (steepness) of AB. The slope of a graph gives the rate of change of the Y-axis quantity with respect to the X-axis quantity, so the slope of a position-time graph is velocity.
5. Velocity-Time Graphs
A velocity-time graph has time on the X-axis and velocity on the Y-axis. NCERT uses three cars on a straight highway:
| Time (s) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| Car 1: steady 72 km/h (m/s) | 20 | 20 | 20 | 20 | 20 | 20 | 20 |
| Car 2: Table 4.5 (m/s) | 0 | 2.5 | 5.0 | 7.5 | 10.0 | 12.5 | 15.0 |
| Car 3: Table 4.6 (m/s) | 15.0 | 12.5 | 10.0 | 7.5 | 5.0 | 2.5 | 0 |
- Car 1: line parallel to the X-axis. Constant velocity, zero acceleration.
- Car 2: straight line sloping up. Velocity rises by equal amounts, so acceleration is constant and along the velocity.
- Car 3: straight line sloping down. Acceleration is constant and opposite to the velocity.
Slope gives acceleration: with A (t1, u) and B (t2, v) on the line, a = (v − u)/(t2 − t1) = BC/CA. Car 2 from 10 s to 20 s: a = (10 − 5)/10 = 0.5 m/s2. Car 3 over the same interval: a = (5 − 10)/10 = −0.5 m/s2.
Area gives displacement: for Car 1 from 0 to 6 s the region under the line is a rectangle of height 20 m/s and width 6 s. Area = velocity × time = 20 × 6 = 120 m, the displacement. In general, the area enclosed by the velocity-time graph and the time axis over an interval equals the displacement in that interval.
For Car 2 from 10 s to 20 s, split the region into a rectangle (5 m/s × 10 s) and a triangle above it (base 10 s, height 5 m/s): displacement = 50 + ½ × 10 × 5 = 50 + 25 = 75 m. As a trapezium: ½ × (5 + 10) × 10 = 75 m, a useful cross-check.
6. Kinematic Equations for Constant Acceleration
With constant acceleration, the acceleration at every instant equals the average acceleration. Let u be the velocity at t = 0, v the velocity at time t, a the acceleration and s the displacement in time t.
First equation: from a = (v − u)/t, at = v − u, so v = u + at (Eq. 4.4a).
Second equation (graphical): the velocity-time graph is a straight line from A on the velocity axis (OA = u) to B at time t (BD = v, where D is on the time axis and OD = t). Draw AC parallel to the time axis, meeting BD at C. Displacement = area OABD:
s = area of rectangle OACD + area of triangle ABC = (OA × OD) + ½ × CA × BC
With OA = u, OD = CA = t and BC = BD − CD = v − u: s = ut + ½ × t × (v − u). Since v − u = at, s = ut + ½ × t × at, so s = ut + ½at2 (Eq. 4.4b).
Third equation (eliminate t): put t = (v − u)/a into Eq. 4.4b:
s = u(v − u)/a + ½a(v − u)2/a2 = (uv − u2)/a + (u2 + v2 − 2uv)/(2a)
s = (2uv − 2u2 + u2 + v2 − 2uv)/(2a) = (v2 − u2)/(2a), so 2as = v2 − u2 and v2 = u2 + 2as (Eq. 4.4c).
Two more (“The Journey Beyond”): region OABD is a trapezium with parallel sides u and v and width t, so s = ½(u + v)t. Putting u = v − at into Eq. 4.4b gives s = (v − at)t + ½at2, so s = vt − ½at2.
| Quantity not given and not asked | Use |
|---|---|
| s | v = u + at |
| v | s = ut + ½at2 |
| t | v2 = u2 + 2as |
| a | s = ½(u + v)t |
Rules: use these only when acceleration is constant. Convert to SI units first. “Starts from rest” means u = 0; “comes to a stop” means v = 0. For motion in both directions, the signs of u, v, a and s give their directions, so fix the positive direction at the start.
Braking (NCERT Example 4.8): brakes give a car a = −4 m/s2. With v = 0, 0 = u2 − 8s, so s = u2/8. At 54 km/h (15 m/s), s = 225/8 = 28.1 m; at 108 km/h (30 m/s), s = 900/8 = 112.5 m. Doubling the speed makes the braking distance four times as long. The real stopping distance also depends on the road surface, the braking capacity of the vehicle and the driver’s reaction time, during which the vehicle keeps moving at its original speed. That is why drivers keep a safe gap from the vehicle ahead and widen it at higher speed. Vehicle-to-vehicle (V2V) communication, being developed in many countries including India, lets vehicles exchange signals and warn drivers of possible collisions.
7. Motion in a Plane and Uniform Circular Motion
Straight-line motion is motion in one dimension. Motion in a plane (a vehicle overtaking another, a kicked ball, a satellite in a circular orbit) is motion in two dimensions. Motion in space, like a bird flying, is three-dimensional.
When an object moves along a circular path, its motion is circular motion. A child on a merry-go-round moving from A to B to C covers the arc ABC as distance, while the displacement is the straight line AC.
- One revolution on a circle of radius R: distance = circumference = 2πR; displacement = 0, so the average velocity is 0.
- If one revolution takes time T: vav = 2πR / T (Eq. 4.5).
- Half a revolution: distance πR, displacement 2R (the diameter).
Uniform circular motion is motion along a circular path with constant (uniform) speed. The speed is the same at every point, so 2πR/T also gives the speed at each point.
Direction changes continuously. An athlete on a rectangular track runs straight along each side and changes direction 4 times per round; on a hexagonal track, 6 times. As the number of sides grows, the turns come more often and the track approaches a circle, where the direction of velocity changes at every point.
Activity 4.5 (marble in a ring): a marble rolls along the inside of a ring placed flat on a smooth surface. When the ring is lifted, the marble moves off in a straight line, in the direction it had at that instant. That direction is along the tangent, a straight line that meets the circle at one and only one point. In circular motion the velocity at any point is along the tangent there, in the direction of motion.
Uniform circular motion is accelerated. Velocity changes if its magnitude, its direction, or both change. Here the direction changes all the time, so the velocity changes and the acceleration is non-zero, even at constant speed. A car going round a bend at a steady 40 km/h is accelerating. Uniform circular motion is an idealised model, and it forms the base for studying planets revolving around the Sun and vehicles taking circular turns.
Formula and Reaction Sheet
| Quantity | Formula | SI unit | When to use it |
|---|---|---|---|
| Average speed | total distance / time interval | m/s | Direction ignored |
| Average velocity | vav = s / t | m/s | Displacement known; state direction |
| Average acceleration | a = (v − u) / (t2 − t1) | m/s2 | Two velocities and the interval |
| First equation | v = u + at | m/s | No s involved |
| Second equation | s = ut + ½at2 | m | No v involved |
| Third equation | v2 = u2 + 2as | - | No t involved |
| Mean-velocity form | s = ½(u + v)t | m | No a involved |
| Braking distance | s = u2 / (2 × magnitude of a) | m | v = 0; grows as u2 |
| Slope of position-time graph | (s2 − s1)/(t2 − t1) | m/s | Gives velocity |
| Slope of velocity-time graph | (v − u)/(t2 − t1) | m/s2 | Gives acceleration |
| Area under velocity-time graph | rectangle + triangle, or trapezium | m | Gives displacement |
| Circular motion, one revolution | distance 2πR; displacement 0; vav = 2πR/T | m; m/s | Radius R, time T |
| Free fall | g = 9.8 m/s2 downward | m/s2 | Dropped objects |
| Unit conversion | km/h × 5/18 = m/s | - | Before any equation |
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Important Definitions
| Term | Definition |
|---|---|
| Position | Distance and direction of an object from the reference point at an instant of time |
| Motion and rest | In motion if the position with respect to the reference point changes with time; at rest if it does not |
| Instant and time interval | An instant is a single clock reading; a time interval is the duration between two instants |
| Total distance travelled | Total length of the path covered; numerical value only; SI unit metre |
| Displacement | Net change in position between two given instants; has magnitude and direction; SI unit metre |
| Scalar and vector | A scalar needs only a numerical value; a vector needs magnitude and direction |
| Average speed | Total distance travelled divided by the time interval |
| Uniform motion | Equal distances in equal intervals of time, for all choices of intervals |
| Average velocity | Displacement divided by the time interval; vav = s/t |
| Rate of change | Change in a quantity divided by the corresponding change in time |
| Average acceleration | Change in velocity divided by the time interval; SI unit m/s2 |
| Acceleration due to gravity (g) | Constant acceleration of a freely falling object, 9.8 m/s2 downward |
| Slope | Steepness of a line; rate of change of the Y-axis quantity with respect to the X-axis quantity |
| Kinematic equations | Equations relating s, t, u, v and a for straight-line motion with constant acceleration |
| Uniform circular motion | Motion along a circular path with constant (uniform) speed |
| Tangent | A straight line that meets a circle at one and only one point |
Solved Examples (NCERT-Based)
Example 1: Athlete on a straight track (NCERT Fig. 4.4)
An athlete starts at O at t = 0 s, reaches A (100 m from O) at t = 10 s, then runs back to B (40 m from O), reaching it at t = 16 s. Find the distance, displacement, average speed and average velocity from t = 0 to t = 16 s.
Solution: Distance = OA + AB = 100 + 60 = 160 m. Displacement = 40 − 0 = 40 m in the positive direction.
Average speed = 160/16 = 10 m/s. Average velocity = 40/16 = 2.5 m/s in the positive direction. From 0 to 10 s she runs one way only, so both come to 100/10 = 10 m/s.
Example 2: A family road trip (NCERT Pause and Ponder)
You drive 200 km north in 3 hours, then 200 km south in 2 hours. Find the average speed and average velocity for the trip.
Solution: Distance = 400 km, time = 5 h. Average speed = 400/5 = 80 km/h. You end where you started, so displacement = 0 and average velocity = 0 km/h. Averaging 66.7 km/h and 100 km/h to get 83.3 km/h is wrong: use total distance over total time.
Example 3: Bus speeding up and braking (NCERT Example 4.3)
A bus at 36 km/h speeds up to 54 km/h in 10 s. Later it brakes and stops in 5 s. Find the average acceleration in both intervals.
Solution: 36 km/h = 36 × 1000/3600 = 10 m/s; 54 km/h = 15 m/s.
(i) a = (15 − 10)/10 = 0.5 m/s2, along the velocity. (ii) a = (0 − 15)/5 = −3 m/s2, opposite to the velocity.
Example 4: Braking distance at two speeds (NCERT Example 4.8)
Brakes give a car an acceleration of −4 m/s2. How far does it travel before stopping if its speed was (i) 54 km/h, (ii) 108 km/h?
Solution: v = 0; (i) u = 15 m/s, (ii) u = 30 m/s. Time is not involved, so use v2 = u2 + 2as: 0 = u2 − 8s, so s = u2/8.
(i) s = 225/8 = 28.1 m. (ii) s = 900/8 = 112.5 m. Double the speed, four times the distance.
Example 5: Motorbike coming to a stop (NCERT Exercise)
A motorbike at 28 m/s with constant acceleration stops after 98 m. Find the acceleration and the time to stop.
Solution: u = 28 m/s, v = 0, s = 98 m. v2 = u2 + 2as: 0 = 784 + 196a, so a = −4 m/s2.
v = u + at: 0 = 28 − 4t, so t = 7 s. Check: s = ½(28 + 0) × 7 = 98 m.
Example 6: Car starting from rest (NCERT Exercise)
A car starts from rest and reaches 24 m/s in 6 s. Find the average acceleration and the distance travelled.
Solution: a = (24 − 0)/6 = 4 m/s2. s = ut + ½at2 = 0 + ½ × 4 × 36 = 72 m.
Example 7: Three stages of a journey (NCERT Exercise)
A car starts from rest, accelerates uniformly to 20 m/s in 5 s, travels at 20 m/s for 10 s, then brakes uniformly to stop in 6 s. Find the total distance.
Solution: Use the area under the velocity-time graph.
- Stage 1 (triangle): ½ × 5 × 20 = 50 m
- Stage 2 (rectangle): 20 × 10 = 200 m
- Stage 3 (triangle): ½ × 6 × 20 = 60 m
Total = 310 m in 21 s.
Example 8: Reaction time and an obstacle (NCERT Exercise)
A bus at 36 km/h sees an obstacle 30 m ahead. The driver takes 0.5 s to react, and the brakes then give a constant acceleration of magnitude 2.5 m/s2. Will the bus stop in time?
Solution: u = 10 m/s. Reaction distance = 10 × 0.5 = 5 m. Braking: 0 = 100 − 2 × 2.5 × s, so s = 20 m. Total = 25 m < 30 m, so the bus stops 5 m short of the obstacle.
Example 9: Tip of a minute hand (NCERT Exercise)
Rohan studies from 6 PM to 7:30 PM. The minute hand of the wall clock is 7 cm long. For its tip, during this interval, find (i) distance travelled, (ii) displacement, (iii) speed, (iv) velocity. Take π = 22/7.
Solution: 90 min = 5400 s = 1.5 revolutions. The tip moves from the 12 mark to the 6 mark.
(i) Distance = 1.5 × 2 × (22/7) × 7 = 66 cm. (ii) Displacement = diameter = 14 cm, from the 12 mark towards the 6 mark. (iii) Speed is constant: 2πR/T = 44 cm / 3600 s = 0.0122 cm/s (the same as 66/5400). (iv) In this chapter, velocity means the velocity at an instant. Its magnitude equals the speed, 0.0122 cm/s, and its direction is along the tangent to the circle at the tip’s position, so the direction changes continuously. Over the whole interval, average velocity = displacement / time = 14/5400 ≈ 0.0026 cm/s, directed from the 12 mark towards the 6 mark.
Competency-Based Questions (with answers)
1. Case-based: Keeping a safe distance
A driving instructor explains that stopping distance = reaction distance (at constant speed while the driver reacts) + braking distance. A learner has a reaction time of 0.8 s and the brakes give an acceleration of magnitude 5 m/s2.
(a) Stopping distance at 72 km/h? u = 20 m/s. Reaction: 20 × 0.8 = 16 m. Braking: 400/10 = 40 m. Total 56 m.
(b) Stopping distance at 36 km/h? u = 10 m/s. Reaction 8 m, braking 100/10 = 10 m. Total 18 m.
(c) Why did halving the speed cut the braking distance to one-fourth but the reaction distance only to one-half? Reaction distance = u × reaction time, proportional to u. Braking distance = u2/(2a), proportional to u2.
2. Data-based: Two cyclists
Cyclists P and Q start together and ride the same way on a straight road.
| Time (s) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| P (m) | 0 | 8 | 16 | 24 | 32 |
| Q (m) | 0 | 2 | 8 | 18 | 32 |
(a) Shape of each position-time graph? P: straight line, uniform motion at 4 m/s. Q: curve getting steeper (2, 6, 10, 14 m in successive 2 s), so Q is accelerating.
(b) Q’s acceleration, if constant? From rest, s = ½at2: 2 = ½ × a × 4, a = 1 m/s2. Check at 8 s: ½ × 64 = 32 m.
(c) Average velocities over 0 to 8 s, and when are their velocities equal? Both 32/8 = 4 m/s. Q’s velocity is v = 0 + 1 × t, which equals 4 m/s at t = 4 s.
3. Source-based: Travellers in the Ganitakaumudi
The Ganitakaumudi (14th century CE) has a problem of two postmen, 210 yojanas apart, walking towards each other at 9 and 5 yojanas per day; they meet after 15 days. A problem in the same style: two messengers start 280 yojanas apart and walk towards each other at 8 and 6 yojanas per day.
(a) When do they meet? They close 14 yojanas a day: 280/14 = 20 days.
(b) How far does each walk? 160 and 120 yojanas; together 280.
4. Assertion-Reason
Assertion (A): An object moving with constant speed can have an acceleration. Reason (R): Velocity changes when the direction of motion changes, even at constant speed.
Answer: (a) Both true, and R correctly explains A. Uniform circular motion is the example.
5. Assertion-Reason
Assertion (A): The magnitude of displacement can be greater than the distance travelled. Reason (R): The magnitude of displacement is the straight-line distance between the initial and final positions.
Answer: (d) A is false, R is true. The straight line is the shortest path, so magnitude of displacement ≤ distance.
6. Assertion-Reason
Assertion (A): The kinematic equations can be used for a car in city traffic whose acceleration keeps changing. Reason (R): They were derived assuming constant acceleration.
Answer: (d) A is false, R is true.
7. Graph reasoning: A lift
A lift moving up has a velocity-time graph that rises in a straight line from 0 to 2 m/s in 2 s, stays at 2 m/s for 6 s, then falls in a straight line to 0 in 2 s.
(a) Acceleration in each stage? 1 m/s2 upward, 0, and −1 m/s2 (1 m/s2 downward, opposite to the velocity).
(b) Height risen? Area = ½ × 2 × 2 + 2 × 6 + ½ × 2 × 2 = 2 + 12 + 2 = 16 m, which is the fourth floor if each floor is 4 m high. Average velocity = 16/10 = 1.6 m/s upward.
8. Competency MCQ: Circular track
An athlete runs one lap of a circular track of radius 49 m at constant speed in 44 s. Which is correct? (A) Average velocity for the lap is 7 m/s. (B) Displacement after half a lap is 154 m. (C) Average speed for the lap is 7 m/s. (D) Acceleration is zero because the speed is constant.
Answer: (C). Distance = 2 × (22/7) × 49 = 308 m, and 308/44 = 7 m/s. (A) fails because displacement for a lap is zero. (B) fails because the half-lap displacement is the diameter, 98 m. (D) fails because the direction of velocity keeps changing.
Important Questions for Exams
1-Mark Questions
- Define displacement. The net change in position of an object between two given instants, with magnitude and direction.
- SI unit of average acceleration? m s−2 (m/s2).
- What does the area under a velocity-time graph give? Displacement in that interval.
- Convert 72 km/h to m/s. 72 × 5/18 = 20 m/s.
- A scooter’s speedometer reading is constant. Can it be accelerating? Yes, when it turns, because the direction of velocity changes.
2-Mark Questions
- My father walked to a shop 250 m away on a straight road, came home for a bag, went to the shop again and returned home. Find distance and displacement. Distance = 4 × 250 = 1000 m. Displacement = 0 (he ends at home).
- A student runs from the ground floor to the fourth floor, then down to the second floor. Each floor is 3 m. Find vertical distance and displacement. 12 m up + 6 m down = 18 m. Displacement = 6 m upward.
- When is (i) average speed equal to the magnitude of average velocity, (ii) average velocity zero while average speed is not? (i) Straight-line motion in one direction. (ii) When the object returns to its starting point.
- “The Earth moves around the Sun.” Can an object kept on the Earth be at rest? Yes, with respect to a reference point on the Earth, since its position relative to that point does not change. With respect to the Sun it moves with the Earth. Rest and motion depend on the reference point.
3-Mark Questions
- A truck slows uniformly from 54 km/h to 36 km/h in 36 s. Find the distance travelled. u = 15 m/s, v = 10 m/s. s = ½(u + v)t = ½ × 25 × 36 = 450 m.
- A car moves at 6 m/s for 2 min, then accelerates at 1 m/s2 for 6 s. Describe the velocity-time graph and find the displacement. A flat line at 6 m/s up to 120 s, then a line rising to 12 m/s at 126 s. Area = 6 × 120 + ½ × (6 + 12) × 6 = 720 + 54 = 774 m.
- Cars A and B start from rest. A reaches 5 m/s in 5 s; B reaches 3 m/s in 10 s. Find their displacements. aA = 1 m/s2, aB = 0.3 m/s2. Triangle areas: sA = ½ × 5 × 5 = 12.5 m; sB = ½ × 10 × 3 = 15 m.
- Explain why uniform circular motion is accelerated, using rectangular, hexagonal and circular tracks. The athlete turns 4 times per round on a rectangle and 6 times on a hexagon. As sides increase the track approaches a circle, where the direction changes at every point. Changing direction means changing velocity, which means acceleration even at constant speed.
5-Mark Questions
- Derive v = u + at and, by the graphical method, s = ut + ½at2. From a = (v − u)/t, v = u + at. On the velocity-time graph from A (0, u) to B (t, v), displacement = area OABD = rectangle OACD + triangle ABC = ut + ½t(v − u). With v − u = at, s = ut + ½at2. Marks go to the labelled graph, the split of the area and the substitution.
- Derive v2 = u2 + 2as. A car at 108 km/h brakes at −6 m/s2. Find the stopping distance and time. Substitute t = (v − u)/a in s = ut + ½at2 and simplify to s = (v2 − u2)/(2a). Numerical: u = 30 m/s, s = 900/12 = 75 m, t = 30/6 = 5 s. Check: ½ × 30 × 5 = 75 m.
- A car’s velocity is 0, 2.5, 5.0, 7.5, 10.0, 12.5, 15.0 m/s at 0, 5, 10 … 30 s. Describe the graph, find the acceleration, and the displacement from 10 s to 20 s and from 0 to 30 s. A straight line through the origin sloping up: constant acceleration. Slope = (10 − 5)/10 = 0.5 m/s2. 10 s to 20 s: 50 + 25 = 75 m. 0 to 30 s: ½ × 30 × 15 = 225 m.
Common Mistakes and Examiner Tips
- Equating distance and displacement. They match only for straight-line motion in one direction. Otherwise work them out separately.
- Displacement without direction. Write “40 m in the positive direction” or “6 m upward”, never just “40 m”.
- Averaging two speeds. Average speed is always total distance / total time.
- Skipping unit conversion. Convert km/h with × 5/18 and minutes to seconds before substituting.
- Losing the minus sign when slowing down. With velocity positive, a is negative while braking; a positive a gives an impossible stopping distance.
- Using the equations when acceleration changes. Split a multi-stage journey into stages, or use the area of the velocity-time graph.
- Mixing up graph rules. Position-time slope = velocity; velocity-time slope = acceleration; velocity-time area = displacement. Read the axis labels first.
- Treating a graph as a route map. A flat position-time line means rest; a flat velocity-time line means constant velocity.
- Forgetting reaction distance. Stopping distance = speed × reaction time + braking distance.
- Zero acceleration in uniform circular motion. The direction of velocity keeps changing, so the motion is accelerated.
- Non-zero average velocity for a full revolution. Displacement is zero after a whole revolution, so average velocity is zero; average speed is 2πR/T.
Quick Revision Points
- Position = distance and direction from a reference point at an instant
- Motion: position changes with time; rest: it does not; both depend on the reference point
- Right of O usually +, left −; keep the choice fixed through a problem
- Distance: path length, no direction; displacement: net change in position, with direction; unit metre
- Magnitude of displacement ≤ distance; equal only for one-direction straight-line motion
- Average speed = distance / time; average velocity = displacement / time; unit m/s
- Uniform motion: equal distances in equal intervals, for all intervals
- 36, 54, 72, 108 km/h = 10, 15, 20, 30 m/s (multiply by 5/18)
- a = (v − u)/(t2 − t1), unit m/s2; negative when slowing with velocity +
- Steady fast motion in a line has zero acceleration
- Free fall: g = 9.8 m/s2 downward
- Position-time: flat = rest, straight = constant velocity, curve = changing velocity; slope = velocity
- Velocity-time: flat = zero acceleration, straight sloping = constant acceleration; slope = acceleration; area = displacement
- v = u + at; s = ut + ½at2; v2 = u2 + 2as; also s = ½(u + v)t
- Kinematic equations hold only for constant acceleration
- Braking distance = u2/(2a), with a the magnitude of the braking acceleration: double the speed, four times the distance
- Stopping distance = reaction distance + braking distance
- One revolution: distance 2πR, displacement 0, vav = 2πR/T
- Uniform circular motion: constant speed, changing direction, so accelerated; velocity along the tangent
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams (periodic tests, half-yearly and annual papers), and the NCERT source gives no fixed mark allocation. It is a numerical chapter, so it usually supplies both definitions and calculations.
| Question type | What this chapter typically supplies |
|---|---|
| MCQ and one-word | SI units, km/h to m/s, reading a graph, zero displacement on a round trip |
| Assertion-Reason | Constant speed with acceleration, displacement versus distance, validity of the equations |
| Short answer (2 to 3 marks) | Distance and displacement, average speed versus velocity, one-step kinematic numericals |
| Long answer (5 marks) | Graphical derivation of the equations, multi-stage journeys, braking with reaction time |
| Case-based | Safe distance on roads, data tables to plot and read, circular tracks and clock hands |
This chapter is the foundation for Class 10 physics: fixing an origin, converting units first and reading slope and area from graphs are used in every later numerical chapter.
Class 9 Science · Chapter 4 – swipe through all 10 cards to understand the whole chapter.
Position and Reference Point
Position is the distance and direction of an object from a fixed reference point at an instant.
Motion means position changes with time; rest means it does not.
- On a line, right of O is usually + and left is −.
- Pick the origin and + direction once, then keep them.
- Rest and motion depend on the chosen reference point.
Distance and Displacement
Distance is the full path length; displacement is the net change in position, with direction.
Both are measured in metre (m).
- O to 100 m and back to 40 m: distance 160 m, displacement 40 m.
- A round trip has zero displacement.
- Equal only for straight-line motion in one direction.
Average Speed and Velocity
Average speed uses total distance; average velocity uses displacement and has a direction.
SI unit m/s; 1 km/h = 5/18 m/s.
- 25 m pool, there and back in 50 s: 1 m/s speed, 0 m/s velocity.
- 200 km N in 3 h, 200 km S in 2 h: speed 80 km/h.
- Uniform motion: equal distances in equal time intervals.
Average Acceleration
Average acceleration is the change in velocity divided by the time interval.
SI unit m/s2; negative when slowing with velocity taken +.
- Bus 10 to 15 m/s in 10 s: a = 0.5 m/s2.
- Bus 15 m/s to rest in 5 s: a = −3 m/s2.
- Steady fast motion in a line has zero acceleration.
Acceleration Due to Gravity
A dropped object gains 9.8 m/s of velocity every second, a constant downward acceleration.
Velocities 0, 9.8, 19.6, 29.4, 39.2 m/s at 0 to 4 s.
- Equal velocity gains in equal times mean constant acceleration.
- Falls 44.1 m in the first 3 s from rest.
- Acceleration points downward, along the motion.
Position-Time Graph
The slope of a position-time graph gives the velocity of the object.
A graph shows position against time, not the route.
- Flat line: object at rest.
- Straight sloping line: constant velocity.
- Upward-bending curve: velocity increasing.
Velocity-Time Graph
Slope gives acceleration; area between the line and the time axis gives displacement.
Split the area into a rectangle and a triangle, or use a trapezium.
- 20 m/s for 6 s: area 120 m.
- 5 to 10 m/s over 10 s: 50 m + 25 m = 75 m.
- Flat line means zero acceleration.
Kinematic Equations
Five quantities s, t, u, v and a are linked when acceleration is constant.
Also s = ½(u + v)t and s = vt − ½at2.
- Use them only for constant acceleration.
- Starts from rest: u = 0; comes to a stop: v = 0.
- Convert km/h to m/s before substituting.
Braking and Safe Distance
Stopping distance is reaction distance plus braking distance.
a = 4 m/s2: 28.1 m at 54 km/h, 112.5 m at 108 km/h.
- Double the speed, four times the braking distance.
- Reaction distance = speed × reaction time.
- Wet roads reduce braking acceleration, so distance grows.
Uniform Circular Motion
Motion on a circle at constant speed, with the direction of velocity changing all the time.
Velocity at each point is along the tangent.
- One revolution: distance 2πR, displacement zero.
- Changing direction means changing velocity, so it is accelerated.
- A marble freed from a ring moves off in a straight line.
📝 Practice Describing Motion Around Us - 10 questions
Concept-check questions · with answers & explanations
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Why A: Step 1 (distance adds every leg): going up 4 floors is 4 × 3 m = 12 m, coming down 2 floors is 2 × 3 m = 6 m, so the total distance is 12 m + 6 m = 18 m. Step 2 (displacement is the net change in position): she ends on the second floor, 2 × 3 m = 6 m above the ground floor where she started, so the displacement is 6 m upward.
Why not B: 12 m is only the upward leg; the 6 m she came down is also distance travelled.
Why not C: 18 m as displacement treats the whole path as if it were all in one direction.
Why not D: 6 m downward is the last leg alone; displacement is measured from the starting point, and she is above it, not below.
Remember: Displacement is measured from where you started to where you stopped, direction included.
Why B: Average speed uses total distance and average velocity uses displacement, both over the same time interval. So their magnitudes are equal exactly when the magnitude of displacement equals the distance travelled. That happens only when the object moves along a straight line in one direction without turning back, because then the path length is the straight-line change in position.
Why not A: At rest both quantities are zero, but that is a trivial case and not the only one; a car driving straight along a highway also satisfies the condition.
Why not C: In a round trip the displacement is zero while the distance is not, so the average velocity is zero and the average speed is not.
Why not D: They are different quantities, but their magnitudes coincide in one-direction straight-line motion, as the textbook notes.
Remember: No turning back means displacement equals distance, so average velocity equals average speed in magnitude.
Why B: Step 1 (unit conversion): u = 54 km/h = 54 × 5/18 = 15 m/s, and v = 0 m/s because the bus stops. Step 2 (average acceleration = (v – u) / t): a = (0 – 15) / 5 = -3 m/s². The minus sign means the acceleration is opposite to the direction of velocity, as it must be when the magnitude of velocity decreases.
Why not A: +3 m/s² has the right magnitude but the wrong sign; a positive acceleration along the motion would make the bus speed up.
Why not C: -10.8 uses 54 km/h directly as if it were in m/s.
Why not D: -0.6 divides by 5 twice (15 / 25); a single division of the change in velocity by the time interval is all that is needed.
Remember: Slowing down in a straight line means acceleration opposite to velocity, so it carries the opposite sign.
Why C: Step 1 (area under the graph = displacement): the region under the sloping line from 10 s to 20 s is a rectangle of height 5 m/s and width 10 s plus a triangle on top of it with base 10 s and height (10 – 5) = 5 m/s. Step 2 (area of rectangle and triangle): rectangle = 5 × 10 = 50 m, triangle = 1/2 × 10 × 5 = 25 m, so displacement = 50 m + 25 m = 75 m. Check: the average of the two velocities, 7.5 m/s, times 10 s also gives 75 m.
Why not A: 50 m is only the rectangle, which assumes the car stayed at 5 m/s for the whole interval.
Why not B: 100 m is the final velocity 10 m/s times 10 s, which assumes the car moved at its final velocity throughout.
Why not D: 25 m is only the triangle, the extra displacement due to the increase in velocity, and leaves out the rectangle.
Remember: Under a sloping velocity-time line, displacement = rectangle + triangle, or average velocity × time.
Why C: Step 1 (unit conversion): u = 36 km/h = 10 m/s. Step 2 (constant velocity during reaction time): in the 0.5 s before braking the bus covers 10 m/s × 0.5 s = 5 m. Step 3 (v² = u² + 2as for braking): 0 = 10² + 2 × (-2.5) × s, so s = 100 / 5 = 20 m. Step 4 (adding the two parts): total stopping distance = 5 m + 20 m = 25 m, which is less than 30 m, so the bus stops 5 m short of the obstacle.
Why not A: 45 m comes from a braking distance of 40 m, obtained by forgetting the factor 2 in 2as, plus the 5 m reaction distance.
Why not B: Nothing in the calculation gives exactly 30 m; this simply assumes the answer must match the gap.
Why not D: 20 m is only the braking distance; the 5 m covered during the reaction time must be added.
Remember: Stopping distance = reaction distance + braking distance; both parts grow with speed.
Why A: The speedometer shows only the magnitude of velocity. If the scooter takes a turn, the direction of its velocity changes even though the reading stays constant. Velocity has both magnitude and direction, so a change in direction alone is a change in velocity, and acceleration is change in velocity per unit time. Hence the scooter is accelerating whenever it turns, and R states exactly why.
Why not B: R is not a separate fact; it is the very reason the assertion holds.
Why not C: R is a correct statement from the chapter: acceleration can result from a change in the magnitude of velocity or its direction, or both.
Why not D: A is true; a vehicle moving at a steady speed round a bend is accelerating, just as in uniform circular motion.
Remember: Turning at constant speed is still acceleration, because direction is part of velocity.
Why C: Step 1 (distance adds every leg): home to shop, shop to home, home to shop, shop to home is four legs of 250 m each, so total distance = 4 × 250 m = 1000 m. Step 2 (displacement is the net change in position): he starts and ends at home, so his position has not changed and the displacement is 0 m.
Why not A: 500 m counts only one round trip out of the two he made, and 250 m treats the shop as the final position.
Why not B: The distance is right but the displacement is measured between the starting and stopping positions, both at home, so it is zero.
Why not D: 750 m counts only three legs; the man made four trips of 250 m each.
Remember: Any journey that ends where it began has zero displacement, however long the path.
Why C: Step 1 (average speed = total distance / total time interval): total distance = 200 km + 200 km = 400 km and total time = 3 h + 2 h = 5 h, so average speed = 400 km / 5 h = 80 km/h. Step 2 (average velocity = displacement / time interval): 200 km north followed by 200 km south brings the family back to the starting point, so displacement = 0 and average velocity = 0 km/h.
Why not A: 80 km/h as velocity treats the 400 km path as displacement; the net change in position is zero.
Why not B: 100 km/h is the speed of the second leg alone (200 km in 2 h), not the average over the whole trip.
Why not D: The two answers are swapped; distance was covered, so the average speed cannot be zero.
Remember: Average speed uses the whole path and the whole time; average velocity uses only the start and finish positions.
Why B: Acceleration depends not on how fast an object moves but on how quickly its velocity changes. The bus keeps the same velocity, so the change in velocity over any interval is zero, and its average acceleration is (v – u)/t = 0. An object can be moving very fast and still have zero acceleration.
Why not A: A high speed by itself does not cause acceleration; only a change in velocity does.
Why not C: 25 m/s is the speed in SI units, and speed has the unit m/s, not m/s²; the numbers describe different quantities.
Why not D: Acceleration is opposite to velocity only when the magnitude of velocity is decreasing, which is not the case here.
Remember: Constant velocity, whatever its size, means zero acceleration.
Why D: Step 1 (average velocity = change in position / change in time): change in position = 80 m – 40 m = 40 m, change in time = 4 s – 2 s = 2 s. Step 2 (slope of the line): v = 40 m / 2 s = 20 m/s. Since the graph is a straight line, this slope is the same at every instant, so the car moves at a constant 20 m/s.
Why not A: 10 m/s divides the change in position 40 m by the final clock reading 4 s instead of by the time interval 2 s.
Why not B: 40 m/s is the change in position in metres, not divided by the time interval.
Why not C: 80 m/s is the final position read straight off the graph, not a velocity.
Remember: On a position-time graph velocity = rise / run = change in position / change in time.
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Frequently Asked Questions
Distance travelled is the total length of the path covered and needs only a number with a unit. Displacement is the net change in position between two instants and needs both magnitude and direction. An athlete who runs 100 m forward and 60 m back has travelled 160 m, but her displacement is 40 m in the forward direction. The magnitude of displacement is never more than the distance.
Average speed is total distance travelled divided by the time interval, and it has no direction. Average velocity is displacement divided by the time interval, and it has the direction of the displacement. A swimmer who swims to the far end of a 25 m pool and back (50 m) in 50 s has an average speed of 1 m/s and an average velocity of 0 m/s.
Yes. Acceleration is the rate of change of velocity, so an object moving at a high but steady velocity in a straight line has zero acceleration. A bus cruising at 90 km/h on a straight highway is an example. Its velocity-time graph is a flat line parallel to the time axis, whose slope is zero.
The slope of a velocity-time graph gives the acceleration, found as the change in velocity divided by the change in time. The area enclosed between the line and the time axis over an interval gives the displacement in that interval. For a car going from 5 m/s to 10 m/s in 10 s, the slope is 0.5 m/s² and the area is 50 m + 25 m = 75 m.
They are v = u + at, s = ut + ½at² and v² = u² + 2as, where u is initial velocity, v is final velocity, a is acceleration, s is displacement and t is time. They apply only to straight-line motion with constant acceleration. Convert all quantities to SI units before substituting, and keep one positive direction throughout.
With v = 0, the equation v² = u² + 2as gives braking distance s = u²/(2a), which depends on the square of the initial speed. Doubling the speed makes the braking distance four times as long. With brakes giving 4 m/s², a car at 54 km/h stops in about 28.1 m, while at 108 km/h it needs 112.5 m. Reaction distance adds to this.
In uniform circular motion the speed stays constant, but the direction of motion changes at every point of the circle. Velocity has direction, so a changing direction means a changing velocity, and any change in velocity means acceleration. That is why a car going round a bend at a steady speedometer reading is still accelerating.
Multiply by 5/18, because 1 km/h = 1000 m divided by 3600 s. So 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s and 108 km/h = 30 m/s. To convert m/s back to km/h, multiply by 18/5. Always convert before using the kinematic equations.
