NCERT Solutions Curiosity Chapter 8 Chapter exercises — Let Us Enhance Our Learning

Book page 118 & 119 Updated on2026-09-05

Q1.

Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.

Answer

Speed = Total distance covered ÷ Total time taken
= 150 m ÷ 10 s
= 15 m/s

Now change m/s into km/h:

150 m = 150 ÷ 1000 km = 0.15 km
10 s = 10 ÷ 3600 h
Speed = 0.15 km ÷ (10/3600) h = 0.15 × 360 = 54 km/h

The short way gives the same answer:

1 m/s = 3.6 km/h
15 m/s = 15 × 3.6 = 54 km/h

Where 3.6 comes from: in 1 second an object at 1 m/s covers 1 m. In 1 hour (3600 s) it covers 3600 m = 3.6 km. So 1 m/s is the same speed as 3.6 km/h — multiply by 3.6 going one way, divide by 3.6 coming back.

Q2.

A runner completes 400 metres in 50 seconds. Another runner completes the same distance in 45 seconds. Who has a greater speed and by how much?

Answer

First runner: speed = 400 m ÷ 50 s = 8 m/s
Second runner: speed = 400 m ÷ 45 s = 8.89 m/s (8 8/9 m/s) ≈ 8.9 m/s

The second runner has the greater speed — the one who took less time for the same distance.

Difference = 8.89 m/s − 8 m/s = 0.89 m/s (exactly 8/9 m/s ≈ 0.9 m/s)

In km/h the numbers come out neatly:

8 m/s = 8 × 3.6 = 28.8 km/h
8.89 m/s = 8.89 × 3.6 = 32 km/h
Difference = 32 − 28.8 = 3.2 km/h

Why less time means more speed: the distance is the same for both, so speed depends only on the time in the denominator. A smaller denominator gives a bigger answer. Cutting 5 s off a 400 m run raised the speed by about 0.9 m/s.

Q3.

A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?

Answer

Total time taken = Total distance covered ÷ Speed

The speed is in m/s, so first put the distance in metre:

360 km = 360 × 1000 m = 3,60,000 m
Time = 3,60,000 m ÷ 25 m/s = 14,400 s
14,400 s ÷ 60 = 240 min = 240 ÷ 60 = 4 h

Check by working in km/h instead:

25 m/s = 25 × 3.6 = 90 km/h
Time = 360 km ÷ 90 km/h = 4 h ✔

Tip: mixing units is the commonest mistake in this chapter. Before dividing, look at the two quantities and ask: 'are they in the same family — m with s, or km with h?'

Q4.

A train travels 180 km in 3 h. Find its speed in: (i) km/h (ii) m/s (iii) What distance will it travel in 4 h if it maintains the same speed throughout the journey?

Answer

(i) In km/h

Speed = 180 km ÷ 3 h = 60 km/h

(ii) In m/s

180 km = 1,80,000 m · 3 h = 3 × 3600 s = 10,800 s
Speed = 1,80,000 m ÷ 10,800 s = 16.67 m/s (that is 50/3 m/s ≈ 16.7 m/s)
Check: 60 ÷ 3.6 = 16.67 m/s ✔

(iii) Distance in 4 h at the same speed

Total distance covered = Speed × Total time taken
= 60 km/h × 4 h = 240 km

Why part (iii) needs the words 'same speed throughout': the formula distance = speed × time uses one single speed for the whole journey. That is allowed only if the motion is uniform — otherwise you would need the average speed for those particular 4 hours, which the question does not give.

Q5.

The fastest galloping horse can reach the speed of approximately 18 m/s. How does this compare to the speed of a train moving at 72 km/h?

Answer

Bring both speeds to the same unit before comparing.

Train: 72 km/h = 72 ÷ 3.6 = 20 m/s
Horse: 18 m/s = 18 × 3.6 = 64.8 km/h

	In m/s	In km/h
Galloping horse	18	64.8
Train	20	72

The train is faster. It moves 2 m/s (that is 7.2 km/h) faster than the fastest galloping horse. Put another way, the horse's speed is 18 ÷ 20 = 0.9, i.e. 90% of the train's speed — remarkably close for an animal.

Why the comparison is impossible without converting: the number 72 looks four times as big as 18, which wrongly suggests the train is four times faster. The units are different, so the raw numbers cannot be compared at all. Always convert first, then compare.

Q6.

Distinguish between uniform and non-uniform motion using the example of a car moving on a straight highway with no traffic and a car moving in city traffic.

Answer

Point of difference	Car on an empty straight highway	Car in city traffic
Type of motion	Uniform linear motion	Non-uniform linear motion
Speed	Stays constant, say 60 km/h, because nothing forces the driver to change it	Changes constantly — slows at red lights and crossings, speeds up when the road clears
Distance in equal intervals of time	Equal: 1 km in every minute	Unequal: 1 km in one minute, then only 300 m in the next
Speedometer reading	Needle stays near one mark	Needle keeps moving up and down, and falls to zero at signals
Distance–time graph	A straight line	A bent, uneven line, flat wherever the car is stopped
Speed at any instant vs average speed	Same as the average speed	Sometimes more, sometimes less than the average speed

Both cars are moving along a straight road, so both are in linear motion. The difference lies only in whether the speed stays constant.

Why real journeys are almost never uniform: even the highway car must slow at a toll plaza or a curve. Uniform linear motion is an idealisation — that is exactly why the book says we have to use average speeds.

Q7.

Data for an object covering distances in different intervals of time are given in the following table. If the object is in uniform motion, fill in the gaps in the table. — Time (s): 0, 10, 20, 30, __, 50, __, 70 ; Distance (m): 0, 8, __, 24, 32, 40, __, 56

Answer

First find the speed from a pair of values that is completely given, then use it for every gap.

Speed = 8 m ÷ 10 s = 0.8 m/s
Check with another complete pair: 24 m ÷ 30 s = 0.8 m/s ✔ and 40 m ÷ 50 s = 0.8 m/s ✔

Now fill the gaps using distance = 0.8 × time and time = distance ÷ 0.8:

Distance at 20 s = 0.8 × 20 = 16 m
Time for 32 m = 32 ÷ 0.8 = 40 s
Seventh column (between 50 s and 70 s) = 60 s, and distance = 0.8 × 60 = 48 m

Time (s)	0	10	20	30	40	50	60	70
Distance (m)	0	8	16	24	32	40	48	56

The values filled in are shown in orange: 16 m, 40 s, 60 s and 48 m.

All eight readings lie on one straight line through the origin — the mark of uniform motion. The orange points are the values filled in.

Why one straight line is enough: in uniform motion the object covers 0.8 m in every second, no matter when you start counting. So distance and time rise together in a fixed ratio, and the graph can only be a straight line through the origin. Any gap can then be read straight off that line.

Q8.

A car covers 60 km in the first hour, 70 km in the second hour, and 50 km in the third hour. Is the motion uniform? Justify your answer. Find the average speed of the car.

Answer

No, the motion is not uniform.

Justification: an object in uniform motion covers equal distances in equal intervals of time. Here the intervals are equal (one hour each) but the distances are not:

1st hour → 60 km · 2nd hour → 70 km · 3rd hour → 50 km
60 km ≠ 70 km ≠ 50 km ⇒ non-uniform motion

Average speed:

Total distance covered = 60 + 70 + 50 = 180 km
Total time taken = 1 + 1 + 1 = 3 h
Average speed = 180 km ÷ 3 h = 60 km/h
In SI units: 60 ÷ 3.6 = 16.67 m/s

Why we add first and divide once: average speed is total distance ÷ total time — never the average of the three separate speeds. Here the two happen to agree ((60+70+50)/3 = 60) only because the three intervals are equal. If the car had covered 60 km in 1 h and 70 km in 2 h, averaging the speeds would give a wrong answer.

Did you know? The car's average speed of 60 km/h is exactly the speed it had in the first hour — yet at no single moment of the second or third hour was it moving 'averagely'. Average speed is a summary of the journey, not a description of any instant in it.

Q9.

Which type of motion is more common in daily life—uniform or non-uniform? Provide three examples from your experience to support your answer.

Answer

Non-uniform motion is far more common in daily life. Uniform linear motion is an idealisation; we seldom find objects moving with a constant speed over long distances or for long intervals of time.

A school bus on its morning route. It speeds up on an open stretch, slows for speed breakers, and stops completely at every pick-up point and red light. In one minute it may cover 600 m, in the next only 100 m.
Walking to school or to the market. You start slowly, walk fast when you are late, slow down at a crossing, and stop to greet a friend. The distance covered in each minute keeps changing.
A cricket ball after it is hit. It leaves the bat very fast, slows down in the air, and slows still more after it hits the ground and rolls to the boundary.

Two more from a train journey: the train of Fig. 8.11 is non-uniform between A and B while gathering speed, and again between C and D while braking — uniform only in the middle stretch B to C.

Why uniform motion is so rare: to keep a speed exactly constant, everything that slows a body down — friction, air resistance, traffic, curves, gradients — must be balanced perfectly, second after second. In everyday situations that almost never happens. The nearest examples are things like the tip of a clock's second hand or a ceiling fan running at a steady setting.

Q10.

Data for the motion of an object are given in the following table. State whether the speed of the object is uniform or non-uniform. Find the average speed. — Time (s): 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 ; Distance (m): 0, 6, 10, 16, 21, 29, 35, 42, 45, 55, 60

Answer

Work out the distance covered in each 10-second interval by subtracting successive distances.

Time (s)	0	10	20	30	40	50	60	70	80	90	100
Distance (m)	0	6	10	16	21	29	35	42	45	55	60
Covered in that interval (m)	—	6	4	6	5	8	6	7	3	10	5

The intervals of time are all equal (10 s each), but the distances covered in them — 6, 4, 6, 5, 8, 6, 7, 3, 10, 5 m — are not equal.

So the speed of the object is non-uniform.

Average speed = Total distance covered ÷ Total time taken
= 60 m ÷ 100 s
= 0.6 m/s
In km/h: 0.6 × 3.6 = 2.16 km/h

The red line joining the readings is uneven — the speed keeps changing. The dashed grey line is the straight path the object would have followed had it moved uniformly at its average speed of 0.6 m/s.

Why the graph is the quicker test: you do not even have to subtract. If the plotted points lie on one straight line, the motion is uniform; if the line bends, it is non-uniform. Here the line is steepest between 80 s and 90 s (10 m in 10 s = 1 m/s, the fastest stretch) and flattest between 70 s and 80 s (3 m in 10 s = 0.3 m/s, the slowest).

Q11.

A vehicle moves along a straight line and covers a distance of 2 km. In the first 500 m, it moves with a speed of 10 m/s and in the next 500 m, it moves with a speed of 5 m/s. With what speed should it move the remaining distance so that the journey is complete in 200 s? What is the average speed of the vehicle for the entire journey?

Answer

Find the time used up in the first two stretches, then see how much time and distance are left.

Total distance = 2 km = 2000 m

First stretch: time = 500 m ÷ 10 m/s = 50 s
Second stretch: time = 500 m ÷ 5 m/s = 100 s
Time used so far = 50 + 100 = 150 s

Remaining distance = 2000 − (500 + 500) = 1000 m
Remaining time = 200 s − 150 s = 50 s
Required speed = 1000 m ÷ 50 s = 20 m/s (= 72 km/h)

Average speed for the whole journey:

Average speed = Total distance ÷ Total time
= 2000 m ÷ 200 s = 10 m/s (= 36 km/h)

The three stretches of the journey. The bar widths are drawn in proportion to the distances.

Why the average is not (10 + 5 + 20) ÷ 3 = 11.67 m/s: the three speeds were held for different lengths of time — 50 s, 100 s and 50 s. The slow 5 m/s stretch lasted twice as long as either of the others, so it pulls the average down. Average speed must always be worked out as total distance ÷ total time.

Check it yourself: add the three times, 50 + 100 + 50 = 200 s ✔, and the three distances, 500 + 500 + 1000 = 2000 m ✔. Both totals agree with the question, so the answer is consistent.

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