NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us (PDF) – 2027

We should keep at least the distance our own vehicle needs to stop — the reaction distance plus the braking distance. At 54 km h⁻¹ on a dry road this is about 43 m.

Taking u = 54 km h⁻¹ = 15 m s⁻¹, reaction time t_r ≈ 1 s and braking |a| = 4 m s⁻²:

Yes — and much more strongly than most people expect, because the braking part grows as the square of the speed.

Doubling the speed does not double the braking distance — it makes it four times as long. The reaction distance (u × t_r) only doubles, so at high speed the braking term dominates completely.

Yes. The total distance travelled and the magnitude of displacement are equal whenever the object moves in one direction along a straight line and never turns back.

Example: between t = 0 s and t = 10 s the athlete of Fig. 4.4 runs only from O to A, without turning back:

Yes. The whole journey — up from O to B and down again from B to O — happens along one and the same vertical straight line.

Reading Fig. 4.5: O is at 0 cm, A at 40 cm, the highest point B at 140 cm, and C at 80 cm — but C is marked on the downward leg, so the ball reaches C only after it has already come down 60 cm from B.

The correct statement is (iii) Its magnitude is less than or equal to the total distance travelled.

• (i) is false — at the last row the ball is back at O and the displacement is exactly 0 cm.
• (ii) is false — at C the displacement is 80 cm while the distance travelled is 200 cm. The displacement can never win, because it is only the straight-line gap between the two end positions.
• (iii) is true — at A both are 40 cm (equal); everywhere after the ball turns back, the displacement is smaller.
• (iv) is false — it says “in all cases”, but at A they are equal, so “less than” does not hold in every case.

The displacement is zero the moment she comes back to her starting point O — because displacement is the net change in position, and her final position is then the same as her initial position.

If she runs from O all the way to A (100 m) and returns to O, then

(i) Total distance travelled.

Justification: fuel is burnt to keep the vehicle moving, against friction and air resistance, over every metre of road it actually covers — that is, over the total distance travelled. Displacement tells us nothing about the path: a trip to a school 4 km away and back has zero displacement, yet it uses fuel for 8 km.

Yes, it is straight-line motion. The track in Fig. 4.6 is a straight (though sloping) line, and the ball stays on it throughout — so the path is a straight line.

Yes, it can be shown on a line like Fig. 4.3. The ball moves along one straight line, so one number — its distance along the track from O — fixes its position. The horizontal line on paper simply stands for the track.

So at A, B, C and D the two are equal.

For the velocity, take north as positive:

(i) When the object moves along a straight line in one direction only, without ever turning back, during that whole time interval.

(ii) When the object returns to its starting position at the end of the interval, so the displacement is zero, while the path length is not.

Examples: Sarang in Example 4.2 swims 50 m up and back in 50 s — average speed 1 m s⁻¹, average velocity 0 m s⁻¹. So does anything completing a full revolution of a circle, or a train that returns to the same platform.

First convert the change in speed into SI units, because the acceleration must come out in m s⁻².

Now look up the “0–100 km h⁻¹” time quoted by the manufacturer or a road test for each car and fill the middle column. Typical published values look like this — treat them as a model; the whole point of the activity is that you find and record real figures yourself.

Worked out for the sample rows:

Put time along the X-axis and position along the Y-axis.

Reason: time is the independent quantity — it goes on steadily whatever the vehicle does — and the position depends on it. By convention the independent quantity is shown on the X-axis. Then the slope of the graph (change in position ÷ change in time) gives the velocity.

A good scale is the one that spreads the data over most of the sheet while keeping the numbers easy to plot. For Table 4.3 the book’s choice is

Check what that uses up: 6 s × 5 divisions = 30 divisions across, and 120 m ÷ 20 m × 5 = 30 divisions up — a comfortable square block on an ordinary sheet.

Plotting (0 s, 0 m), (1 s, 20 m), (2 s, 40 m) … (6 s, 120 m) and joining them gives a single straight line through the origin.

The shape tells you at once whether the velocity is constant or changing, because the slope of a position–time graph is the velocity.

Three things can be pulled straight off it:

1. The position of the object at any instant — read the Y-value directly above that time.
2. The displacement between any two instants — subtract the two Y-values, s₂ − s₁.
3. The average velocity between those instants — the slope of the line joining the two points.

Take A at t₁ = 2 s (position s₁ = 40 m) and B at t₂ = 4 s (position s₂ = 80 m) on the straight line of Fig. 4.11c.

From A, the line parallel to the Y-axis drops to t₁ on the time axis; the line parallel to the X-axis runs across to s₁ on the position axis. Doing the same from B marks off t₂ and s₂.

The right-angled triangle ABC has the graph line AB as its hypotenuse. Its two perpendicular sides are:

• CA — the horizontal side — represents the change in time, t₂ − t₁.
• BC — the vertical side — represents the change in position, s₂ − s₁, which is the displacement between those two instants.

Yes — dividing the two sides of the triangle reproduces the definition of average velocity exactly.

Substituting the four values read off Fig. 4.14:

Here the slope is the acceleration, so the shape reports on how the velocity is changing.

Three quantities can be obtained from a velocity–time graph:

1. The velocity at any instant — read the Y-value directly.
2. The average acceleration — the slope of the line between two points.
3. The displacement — the area enclosed between the line and the time axis for that interval.

Yes — the displacement, from the area enclosed between the graph line and the time axis.

For the accelerating car of Fig. 4.18b, the displacement between 10 s and 20 s is the area of the shape ABDE, which splits into a rectangle and a triangle:

Yes. A vehicle cannot stop where the driver decides to stop — it stops where physics allows, and that place is set by u², not by u.

Using the chapter’s own braking figure of |a| = 4 m s⁻² and a reaction time t_r = 1 s:

Going from 54 to 108 km h⁻¹ doubles the speed but makes the stopping distance more than three times longer — because the braking part alone becomes four times longer.

The child on the merry-go-round moves from A to B to C along the rim (Fig. 4.22).

• Distance travelled = the length of the curved arc ABC actually followed along the circle.
• Displacement = the straight line AC, joining the starting and finishing positions, directed from A towards C.

The two are not equal: the arc bulges outward while the chord cuts straight across, so arc ABC > chord AC.

One full revolution means going all the way round the rim once, so the distance travelled is the circumference of the circular path.

The displacement, on the other hand, is zero, because the child ends the revolution at exactly the same position from which it began.

Yes, the direction changes continuously — at every single instant. The velocity at a point is directed along the tangent to the circle at that point, in the direction of motion, and the tangent turns as the object goes round.

Since the velocity keeps changing, uniform circular motion is accelerated motion, even though the speed stays constant.

Use a stiff, fairly wide ring — the empty cardboard core of an adhesive-tape roll is ideal — and one smooth glass marble that rolls freely inside it.

Rest the ring flat on a smooth table or floor and flick the marble in along the inner wall, not towards the centre, so that it hugs the boundary and circles round.

Prediction: the marble will stop curving and will roll away in a straight line — along the tangent to the circle at the point where it happened to be when the ring left it.

It will not keep circling, and it will not fly outward along a radius from the centre. Write your prediction down before you test it.

Observation: the marble immediately leaves the circle and rolls off in a straight line. It does not continue in circular motion.

The straight line it takes is the tangent to the circle at the exact point where the marble was when the ring was lifted — so if you repeat the activity and lift the ring at a different point, the marble goes off in a different direction.

Reason: once released, the marble simply continues to move in the direction in which it was moving at that instant; nothing is left to bend its path.

Do it at least four or five times, lifting the ring at a clearly different point of the circle each time, and mark or note the direction the marble takes away.

Every trial should give the same finding: a straight-line escape, always along the tangent at the release point.

Break the trip into its four legs. Each leg is 250 m long.

For the displacement, take home as the origin and the direction of the shop as positive:

Height of one floor = 3 m. Measure heights above the ground floor, taking upward as positive.

(i) Total vertical distance travelled

(ii) Displacement from the starting point

Yes, it certainly can be accelerating — if she is changing direction, for example going round a bend or a roundabout.

Reason: acceleration is the rate of change of velocity, and velocity has both magnitude and direction. The speedometer shows only the magnitude (speed). A change in direction alone therefore gives an acceleration — as in uniform circular motion.

The only way her scooter can have zero acceleration is if it moves along a straight road with that constant speedometer reading.

Average acceleration — use Eq. (4.3b):

Distance travelled — taking the acceleration to be constant, use Eq. (4.4b):

Acceleration — use Eq. (4.4c):

The minus sign says the acceleration is directed opposite to the velocity — the motorbike is slowing down.

Time taken — now use Eq. (4.4a):

No. Their velocities are never equal.

Justification, in two steps:

1. Both graphs are straight lines. A straight position–time graph means the slope never changes, so each object moves with its own constant velocity throughout.
2. The line for A is steeper than the line for B at every point. Since the slope of a position–time graph is the velocity, v_A > v_B — and since both are constant, that inequality holds for the whole journey.

The correct options are (i) and (ii).

(i) Correct. Same initial and final positions means the same displacement, over the same 10 s.

(ii) Correct. Because neither object reverses, the distance travelled equals the magnitude of the displacement for each of them — and those are equal.

(iii) Wrong — A does not cover a shorter distance; both cover the same distance in the 10 s.

(iv) Wrong — although B is slower than A in the early part of the interval, it is correspondingly faster in the later part, and the two effects cancel over the full 10 s.

First convert both speeds to SI units, because the time is in seconds.

With constant acceleration the average velocity is the plain mean of u and v, so:

Phase 1 (0 s – 5 s), speeding up:

Phase 2 (5 s – 15 s), constant velocity:

Phase 3 (15 s – 21 s), braking:

Yes — the bus stops 5 m short of the obstacle.

Step 1 — reaction distance (no braking for 0.5 s):

Step 2 — braking distance (u = 10 m s⁻¹, v = 0, a = − 2.5 m s⁻²), using Eq. (4.4c):

Step 3 — total stopping distance.

Both statements can be true at the same time, because rest and motion are always stated with respect to a chosen reference point.

Conclusion: an object kept on the Earth is at rest with respect to the Earth, and in motion with respect to the Sun. In everyday life we silently take the Earth as the reference point, which is why we call a book on a table “at rest”.

(i) Constant velocity (20 s – 100 s) — shade the rectangle (blue):

(ii) Velocity decreasing (100 s – 120 s) — shade the trapezium (rose):

Total displacement (0 s – 120 s):

Average acceleration (0 s – 120 s):

She runs in one direction on a straight road, so the distance she ran = area under the velocity–time graph. Points read from Fig. 4.31:

Area, strip by strip (area of a trapezium for each strip):

Rectangle (0 s – 120 s):

Trapezium (120 s – 126 s):

Accelerations (both start from rest, so u = 0):

Velocities at five instants (v = u + at = at):

Displacements = areas of the two triangles:

R = 7 cm = 0.07 m. From 6:00 PM to 7:30 PM = 90 min = 5400 s, in which the minute hand makes 90/60 = 1.5 revolutions.

(i) Distance travelled (1.5 revolutions):

(ii) Displacement (the tip ends at the 6 mark, diametrically opposite its start):

(iii) Speed (average speed = distance ÷ time):

(iv) Velocity (average velocity = displacement ÷ time):

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 take the same time T for one turn, but the numbers move on a bigger circle.

Why the numbers fade: in the short time for which the eye holds an image, a fast-moving number sweeps a longer arc than a letter, so its image smears more and merges with the next number.

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.

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.

Derivation 1 — s = vt − ½at²

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 trapezium OABD, whose parallel sides are OA = u and DB = v, and the distance between them is OD = t.

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:

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.

Poster idea: on a dry road (|a| = 4 m s⁻², reaction time 1 s) a vehicle at 36 km h⁻¹ needs about 22.5 m to stop, but at 72 km h⁻¹ it needs about 70 m — double the speed, more than three times the stopping distance. Skit idea: two friends riding home, one keeping a 3-second gap and one tailgating; a sudden brake ahead shows the difference.