NCERT Class 9 Science Chapter 4 Activities — Describing Motion Around Us
NCERT Class 9 Science — Describing Motion Around Us, Activities 4.1 to 4.5 on pages 51 to 67.
Also here. Solutions to the end-of-chapter questions for this chapter and the in-text questions are on their own pages.
Five activities, and only one of them involves apparatus. This is a chapter about describing motion, so its activities are mostly about extracting meaning from a table, a graph or a figure — which makes them easy to set but easy to rush.
They map onto the chapter’s argument in order: 4.1 separates distance from displacement, 4.2 puts a number on acceleration, 4.3 and 4.4 build and then read a position–time graph, and 4.5 introduces circular motion, where velocity changes while speed does not.
Key insight. The whole chapter turns on one distinction that Activity 4.1 makes concrete: distance is how far you went, displacement is where you ended up. Every later idea inherits it — speed uses distance, velocity uses displacement, and the reason a body in uniform circular motion is accelerating even at constant speed is that its velocity keeps changing direction.
Activity 4.1
Fig. 4.5 shows a ball thrown vertically upwards from O. It rises in a straight line to B and falls back to O. Can this be called motion in a straight line? Fill in Table 4.1 with the total distance travelled and the displacement from O at each of the positions O, A, B, C and O. Then decide which statement about displacement is true.
What you should observe. Yes, it is motion in a straight line — the ball goes up and comes back along the same line. The two dotted lines in the figure are drawn apart only so the upward and downward journeys can be told apart.
| Position | Total distance travelled from O | Displacement from O |
|---|---|---|
| O | 0 cm | 0 cm |
| A | 40 cm | 40 cm upward |
| B | 140 cm | 140 cm upward |
| C | 200 cm | 80 cm upward |
| O | 280 cm | 0 cm |
The true statement is (iii): the magnitude of displacement is less than or equal to the total distance travelled.
Where the numbers come from. Fig. 4.5 marks the heights: A at 40 cm, B at 140 cm, C at 80 cm. On the way down to C the ball has already travelled 140 cm up and then 140 − 80 = 60 cm down, so 200 cm in total, while its displacement is only 80 cm. Back at O it has travelled 140 + 140 = 280 cm with zero displacement.
Why the last row is the whole point. The ball ends exactly where it started. Distance 280 cm, displacement zero. Nothing else in the chapter makes the difference between the two quantities as sharp as a journey that covers nearly three metres and gets nowhere.
Why (i), (ii) and (iv) are wrong. (i) fails at the last row — displacement is zero there. (ii) is impossible: the straight-line distance between two points can never exceed the path length between them. (iv) says “less than in all cases”, but at A and B the two are equal, because the ball has not turned round yet. The word to argue over is “less than” against “less than or equal to”.
What goes wrong. Filling in C’s distance as 80 cm, by reading the height rather than adding up the journey. Distance accumulates and never decreases; displacement is measured from O each time. Making students state which of the two they are computing before writing each cell fixes it.
Activity 4.2
The magnitude of a car’s average acceleration is usually quoted as the time it takes to go from 0 to 100 km/h. Look up that time for several cars, record them in Table 4.2, and calculate the magnitude of the average acceleration for each.
$$a_{av} = \frac{\Delta v}{\Delta t}, \qquad 100\ \text{km h}^{-1} = \frac{100 \times 1000}{3600}\ \text{m s}^{-1} = 27.8\ \text{m s}^{-1}$$
So a car quoted at 10 s has $a_{av} = 27.8/10 \approx$ 2.8 m s⁻²; one at 5 s has about 5.6 m s⁻²; a sports car at 3 s, about 9.3 m s⁻² — close to g.
Why the unit conversion is the real exercise. The figures are advertised in km/h and seconds; acceleration is wanted in m s⁻². Dividing 100 by the time gives a number in km h⁻¹ s⁻¹, which is not the answer. The 3.6 factor is where nearly every wrong answer comes from.
Why the comparison is worth making. A sports car at 0–100 in 3 seconds accelerates at roughly the same rate as a dropped stone. That is a genuinely surprising thing to be able to say, and it makes an abstract number concrete.
What goes wrong. Treating the quoted time as though the car accelerated uniformly throughout. It does not — this is an average acceleration over the interval, which is exactly what the chapter is defining, and the distinction is worth stating aloud.
Activity 4.3
Take a sheet of graph paper. Draw two perpendicular lines meeting at the origin O; mark the horizontal one OX and the vertical one OY. Using Table 4.3, decide which quantity goes on which axis, choose a scale, plot the points, and join them.
What you should observe. Table 4.3’s positions — 0, 20, 40, 60, 80, 100, 120 m at 0 to 6 s — fall on a perfectly straight line through the origin.
Time on the X-axis, position on the Y-axis. The points lie on a straight line, so the vehicle is moving with constant velocity — 20 m every second.
Why time goes on the X-axis. By convention the independent variable — the one you choose or simply let run — goes horizontally, and the quantity that depends on it goes vertically. Position depends on time, not the other way round.
Why the scale matters as much as the points. A scale chosen badly squashes the graph into a corner and makes the slope impossible to measure in Activity 4.4. The plotted region should fill most of the paper.
What goes wrong. Joining the points with a series of short line segments rather than recognising a single straight line, or forcing a line through points that genuinely curve. The shape is the result — the chapter’s whole next section reads meaning from it.
Activity 4.4
On the position–time graph plotted in Activity 4.3, take a part of the line — say AB — as in Fig. 4.14. From A draw a line parallel to the X-axis and another parallel to the Y-axis, forming a right triangle. Read off the times and positions, and calculate.
What you should observe. The ratio you compute is the velocity, and it is the same wherever on the line you take the triangle.
$$v_{av} = \frac{\text{BC}}{\text{CA}} = \frac{s_2 – s_1}{t_2 – t_1} = \frac{80\ \text{m} – 40\ \text{m}}{4\ \text{s} – 2\ \text{s}} = \frac{40\ \text{m}}{2\ \text{s}} = 20\ \text{m s}^{-1}$$
This ratio is the slope of the line, and the slope of a position–time graph is the velocity.
Why the triangle is drawn rather than the values just read off. It makes the subtraction visible. BC is the change in position and CA the change in time, and seeing them as the two sides of a triangle is what makes “slope” mean something rather than being a formula.
What the chapter generalises it to. “The slope of a graph gives information about the rate of change of the quantity shown on Y-axis with respect to the quantity shown on X-axis.” That sentence is worth keeping, because the same reasoning gives acceleration from a velocity–time graph later.
Why a straight line means constant velocity. Take the triangle anywhere along it and the ratio comes out the same. A curved graph gives a different slope at each point, which is exactly what changing velocity means.
What goes wrong. Reading the slope off the picture rather than the axes — a line that looks like 45° on the paper is only 1 m/s if the scales happen to match, and they usually do not. Always take the numbers from the axes.
Activity 4.5
Take a ring — an adhesive tape ring will do — and a marble. Lay the ring flat on a smooth surface and roll the marble inside it so that it runs around the inner boundary. Predict what will happen if you lift the ring while the marble is moving. Then, after one or two revolutions, lift it and watch.
What you should observe. The marble does not carry on curving, and it does not fly outwards either.
The moment the ring is lifted, the marble travels off in a straight line along the tangent to the circle at the point where it was released.
Why this is the chapter’s point about velocity. In uniform circular motion the speed is constant but the direction changes continuously, so the velocity is changing — and the body is accelerating even though it is not speeding up. The ring was providing the inward force that kept bending the path. Remove it, and with no force left to turn the marble it goes straight, which is the direction its velocity had at that instant: the tangent.
Why “tangent” and not “outwards”. Students overwhelmingly predict that the marble flies radially outwards, away from the centre. It does not. Its velocity was always along the tangent, and with nothing pushing it, that is the direction it keeps. The prediction step in the instructions is there precisely so the class commits to an answer before finding out.
Where the chapter puts this. Immediately after the argument about a runner on a rectangular track, then a hexagonal one, then a track with more and more sides — which approaches a circle, with the direction of velocity changing continuously rather than at a few corners. The chapter is honest that “uniform circular motion is an idealised model of the real-world situations”, but a useful one.
What goes wrong. A surface that is not smooth, or a ring lifted slowly. The marble must be moving freely and the ring must come away cleanly and quickly, or friction stops the marble before the straight-line path is visible. It is worth having someone watch from directly above.
Common mistakes
- Reading Activity 4.1’s distance column off the height scale. Distance accumulates over the whole journey; displacement is measured from O. At C they are 200 cm and 80 cm, and confusing the two is the commonest error in the chapter.
- Claiming displacement can never be zero. The ball ends at O with zero displacement after travelling 280 cm. That single row disposes of option (i).
- Forgetting the 3.6 in Activity 4.2. 100 km/h is 27.8 m/s, not 100. Every acceleration computed without the conversion is wrong by that factor.
- Putting position on the X-axis in Activity 4.3. Time is the independent variable and goes horizontally, which is also what makes the slope come out as a velocity rather than its reciprocal.
- Judging the slope in Activity 4.4 by eye. The apparent steepness depends on the scales chosen. Read the numbers off the axes.
- Predicting that the marble in Activity 4.5 flies outwards. It leaves along the tangent. This is the single most valuable misconception in the chapter to expose, which is why the activity asks for a prediction first.
Practise next
- The end-of-chapter questions, Revise, Reflect, Refine, where the distance-against-displacement distinction from Activity 4.1 and the graph slopes from Activity 4.4 are used on unfamiliar journeys.
- The in-text Pause and Ponder questions of this chapter, five across two boxes, which test the same distinction in shorter form.
- Chapter 6, How Forces Affect Motion, which answers the question Activity 4.5 raises: what was the ring actually doing to the marble, and what happens when that force is removed.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.