NCERT Solutions for Class 9th Science Chapter 4 Projects and extension work — The Journey Beyond

Book page 71 Updated on2026-09-08

Q1.

Take a cardboard disc (radius ~ 8 cm) (Fig. 4.33). Write numbers 1 to 12 on the outer part (7 cm from the centre) and the letters ‘ABCDEF’ on the inner part (4 cm from the centre), using the same font size. Spin the disc slowly, then faster, and observe how the numbers and letters appear. Why do the numbers fade or disappear while the letters remain visible? Are the speeds of the numbers and letters the same or different?

Answer

What you will see: spun slowly, both rings stay readable. Spun fast, the outer numbers smear into a grey blur first, while the inner letters can still be made out.

Are the speeds the same? Different. Both rings complete one revolution in the same time T — the disc is rigid, so nothing can lag behind — but they travel on circles of different radius.

speed of the numbers = 2π × 7 cm / T
speed of the letters = 2π × 4 cm / T
ratio = 7 : 4 = 1.75
so the numbers move 1.75 times as fast as the letters

Put in numbers: if the disc turns once in 0.5 s,

numbers: v = 2 × 3.14 × 0.07 m / 0.5 s = 0.88 m s⁻¹
letters: v = 2 × 3.14 × 0.04 m / 0.5 s = 0.50 m s⁻¹

Why the numbers fade: the eye holds an image for about 1/16 s. In that time a number sweeps a much longer arc than a letter does, so its image is smeared over a longer stretch of the retina. Beyond a certain speed each number's smear overlaps the next one's, and the ring dissolves into a uniform grey. The slower letters, smeared over a shorter arc, stay separated for longer.

Why it happens: what is genuinely equal for the two rings is the number of turns per second — the rate of turning. What is not equal is the linear speed, because speed = (arc length)/(time) and the outer arc is longer for the same turn. This is the same reason the tip of a fan blade whistles while its root does not, and why the outer horses on a merry-go-round feel faster than the inner ones.

Q2.

Many smartphones have an inbuilt accelerometer that can detect very small accelerations. Install an app, such as Phyphox (phyphox.org) and open ‘Accelerometer (without g)’. Note the readings when (i) the phone is on an outstretched palm, and (ii) the phone is kept on the floor. What differences do you observe? What does this tell you about motion and acceleration in real situations? (Such tiny, involuntary movements are also studied in medical research, for example, in movement disorders) This activity is recommended to be performed as a classroom group activity facilitated by teacher.

Answer

How to do it: open Phyphox, choose Accelerometer (without g) so that the steady 9.8 m s⁻² of gravity is already subtracted, and let the graph run for about 30 s in each case without touching the screen.

Where the phone is	Typical reading	What the trace looks like
(i) On an outstretched palm	roughly 0.05 – 0.5 m s⁻²	Constantly wobbling, never settling; small irregular spikes
(ii) On the floor	close to 0, about 0.01 m s⁻² or less	Almost a flat line — only sensor noise, plus a jump when someone walks past

What this tells us:

A hand is never truly still. Muscles hold it up by continuously making tiny corrections, and each correction is a small change of velocity — that is, a real acceleration.
"At rest" is an idealisation, exactly like the idealised motions of Section 4.1. Real objects sit close to rest, not exactly at rest.
The floor reading is not exactly zero either; that residue is instrument noise, and it sets the smallest acceleration the phone can honestly report.

Why it happens: an accelerometer responds to change of velocity, not to velocity itself. That is why a phone lying still on the floor reads nearly zero while a phone in a car moving smoothly at 60 km h⁻¹ also reads nearly zero — and why the tiny tremor of a held hand shows up so clearly. This sensitivity to small accelerations is exactly what makes such sensors useful in medicine: doctors use them to record involuntary tremor in movement disorders, where the size and frequency of the wobble carry diagnostic information.

Try This: hold the phone at arm's length instead of close in, and the readings grow — a longer lever arm magnifies every small twitch at the shoulder.

Q3.

For motion in a straight line with constant acceleration, we derived two primary equations given by Eq. (4.4a) and (4.4b). Using these two equations, three more equations can be derived, out of which we derived one given in Eq. (4.4c). Derive the remaining two equations given below: s = vt − ½at² and s = ½(u + v)t. In mathematics, you have learnt the formula for calculating the area of a trapezium. Using that formula, derive the second equation given above.

Answer

Derivation 1 — s = vt − ½at²

Start from the first primary equation and make u the subject:

v = u + at [Eq. 4.4a]
⇒ u = v − at

Substitute this u into the second primary equation:

s = ut + ½at² [Eq. 4.4b]
s = (v − at)t + ½at²
s = vt − at² + ½at²
s = vt − ½at² proved

Why it happens: compare it with Eq. (4.4b). That equation measures the displacement forward from the start, adding the ½at² the acceleration contributes. This one measures backward from the end: had the object kept its final velocity v for the whole time it would have covered vt, but it was slower than v earlier on, so ½at² has to be taken away.

Derivation 2 — s = ½(u + v)t, using the area of a trapezium

In the velocity–time graph of Fig. 4.19 the displacement is the area of the figure OABD. That figure is a trapezium: OA and DB are its two parallel sides (both vertical), and OD is the perpendicular distance between them.

The shaded trapezium OABD: parallel sides u and v, height t. Its area is the displacement.

area of a trapezium = ½ × (sum of the parallel sides) × (distance between them)

here: parallel sides are AO = u and BD = v; distance between them is OD = t

s = area of OABD = ½ × (u + v) × t
s = ½(u + v)t proved

Check it yourself, algebraically: add the two equations s = ut + ½at² and s = vt − ½at². The ½at² terms cancel:
2s = ut + vt = (u + v)t ⇒ s = ½(u + v)t ✓

Why it happens: ½(u + v) is just the plain average of the starting and finishing velocities, and s = (average velocity) × time. That plain average is correct only because the acceleration is constant, which makes the velocity–time graph a straight line — and a straight line spends as much time above its mid-value as below it. For a curved graph the mid-value would not be the average, and this equation would fail.

Q4.

Plot graphs for data given in Table 4.4, using different X and Y scales, on different graph papers. Compare the graphs to find how the appearance of graph is affected by the choice of scales and decide which scale is better and why. Now repeat this with any graph plotting app. Such apps generally automatically adjust the axes to fit the data well on the screen.

Answer

Method. Table 4.4 runs from (0 s, 0 m) to (12 s, 36 m). Plot it three times on separate sheets, changing only the scales:

Sheet	X-axis scale	Y-axis scale	How the curve looks
1 (the book's)	5 divisions = 2 s	5 divisions = 5 m	Curve fills the sheet; the bend is clearly visible
2	5 divisions = 2 s	5 divisions = 20 m	Squashed flat — looks almost like a straight line near the bottom
3	5 divisions = 5 s	5 divisions = 5 m	Squeezed sideways — looks far steeper than it is

What to conclude. Sheet 1 is the better choice, for two reasons:

It uses most of the available paper, so each point can be plotted and read to the nearest small division — the best accuracy the sheet allows.
It keeps the curvature visible. On sheet 2 the curve is so flattened that a student could easily mistake it for a straight line and wrongly conclude that the vehicle moves with constant velocity.

Why it happens: the scale changes the picture, never the physics. A slope measured off sheet 3 looks steeper, but when you divide the change in position by the change in time using each sheet's own scale, you get exactly the same velocity. What a poor scale really costs you is precision (points crammed together cannot be read finely) and honest interpretation (a squashed curve can hide its own shape).

Tip: when you repeat this in a plotting app, notice that the app picks its axis ranges from the data itself — roughly the sheet 1 choice. That is the rule to imitate by hand: choose the scale after looking at the largest and smallest values you have to fit.

Q5.

Talk to a motor mechanic about how a vehicle's braking or stopping distance is affected by: (i) wet roads, (ii) worn-out tyres, (iii) higher vehicle mass, (iv) driving at night, (v) fog, (vi) severe weather (rain, snow, storm), and (vii) driver reaction time. Using this information, design safety posters for your school and prepare a short skit to present it in the assembly.

Answer

Method. Take the two-term formula from this chapter as your interview framework, and ask the mechanic which term each factor attacks:

stopping distance = u t_r + u² / (2|a|)
(reaction) (braking)

A good write-up should record, for each factor: what the mechanic says happens, which term it changes, and roughly by how much.

Sample answer:

Factor	Which term it affects	Effect on stopping distance
(i) Wet road	Lowers \|a\| (less grip between tyre and road)	Braking distance can roughly double
(ii) Worn-out tyres	Lowers \|a\| — bald treads cannot channel water away	Much longer, and far worse in the wet
(iii) Higher vehicle mass	Lowers \|a\| for the same braking force	A loaded truck needs a longer distance than an empty one
(iv) Driving at night	Raises t_r — the hazard is seen later	Longer reaction distance; also less time to react at all
(v) Fog	Raises t_r sharply and often lowers \|a\| (damp road)	Both terms grow — the reason for very low fog speed limits
(vi) Rain, snow, storm	Lowers \|a\| a great deal	Braking distance can become several times longer
(vii) Driver reaction time	The t_r term directly	Every extra second at 15 m s⁻¹ adds 15 m before braking even starts

For the poster, put one calculation on it rather than a slogan — numbers persuade. For example, with |a| = 4 m s⁻² and t_r = 1 s on a dry road:

at 36 km h⁻¹ (10 m s⁻¹): 10 m + 12.5 m = 22.5 m
at 72 km h⁻¹ (20 m s⁻¹): 20 m + 50 m = 70 m
double the speed → more than three times the stopping distance

For the skit, a simple structure works: two friends riding home, one keeping a 3-second gap and one tailgating; a sudden brake ahead; freeze the action and have a third student read out the two stopping-distance calculations before showing the two outcomes. End with the rule the audience should carry away — keep a time gap, not a fixed gap, and increase it in rain or fog.

Why it happens: notice that six of the seven factors do not change the driver's speed at all — they change how far the vehicle needs to stop at that speed. That is the message worth putting on the poster: a speed that is safe on a dry, clear road can be dangerous on the very same road in rain, with the very same vehicle and driver.

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