Measurement of Time and Motion Class 7 (Curiosity)

Q 1
Question
Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.
🔗 Concept:
8.3
Speed
Speed = 150 m ÷ 10 s — express the answer in km/h.
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Speed = 15 m/s = 54 km/h
Explanation
📋 Given
Distance = 150 m
Time = 10 s
🔄 Unit Conversion
Units are already m and s → answer in m/s first.
To convert m/s to km/h: multiply by 3.6.
Why: 1 km = 1000 m & 1 h = 3600 s, so 1 km/h = 1000/3600 m/s = 1/3.6 m/s.
📐 Formula
Speed = Distance ÷ Time
Step 1 — Find speed in m/s
$$\text{Speed} = \frac{150 \text{ m}}{10 \text{ s}} = 15 \text{ m/s}$$
Step 2 — Convert to km/h
Multiply by 3.6 (since 1 m/s = 3.6 km/h).
$$15 \times 3.6 = \boxed{54 \text{ km/h}}$$
Speed = 15 m/s = 54 km/h — multiply m/s by 3.6 to convert.

Q 2
Question
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?
🔗 Concept:
8.3
Speed
Two runners, 400 m — who is faster and by how much?
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Runner B (45 s) is faster by ≈ 0.89 m/s
Explanation
📋 Given
Runner A: distance = 400 m, time = 50 s
Runner B: distance = 400 m, time = 45 s
🔄 Unit Conversion
No conversion needed — distance in m, time in s → speed in m/s.
📐 Formula
Speed = Distance ÷ Time
Step 1 — Speed of Runner A
$$\text{Speed}_A = \frac{400 \text{ m}}{50 \text{ s}} = 8 \text{ m/s}$$
Step 2 — Speed of Runner B
$$\text{Speed}_B = \frac{400 \text{ m}}{45 \text{ s}} \approx 8.89 \text{ m/s}$$
Step 3 — Compare
Runner B has a higher speed (less time for same distance).
$$\text{Difference} = 8.89 - 8 = \boxed{0.89 \text{ m/s}}$$
Runner B is faster — 8.89 m/s vs 8 m/s, difference ≈ 0.89 m/s.

Q 3
Question
A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?
🔗 Concept:
8.3.1
Speed, Distance, and Time
Train at 25 m/s, distance 360 km — find time taken.
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Time = 14,400 s = 4 h
Explanation
📋 Given
Speed = 25 m/s
Distance = 360 km
🔄 Unit Conversion
Speed is in m/s. Distance must also be in metres.
360 km → multiply by 1000 → 360,000 m.
Why: 1 km = 1000 m.
📐 Formula
Time = Distance ÷ Speed
Step 1 — Convert distance to metres
360 km × 1000 = 360,000 m
Step 2 — Calculate time in seconds
$$\text{Time} = \frac{360{,}000 \text{ m}}{25 \text{ m/s}} = 14{,}400 \text{ s}$$
Step 3 — Convert seconds to hours
Divide by 3600 (since 1 h = 3600 s).
$$\frac{14{,}400}{3600} = \boxed{4 \text{ h}}$$
Time = 14,400 s = 4 h — convert distance to same unit as speed before using the formula.

Q 4
Question
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?
🔗 Concepts:
8.3
Speed
|
8.3.1
Speed–Distance–Time
Train: 180 km in 3 h — find speed in km/h, m/s, and distance in 4 h.
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(i) 60 km/h (ii) ≈ 16.67 m/s (iii) 240 km
Explanation
📋 Given
Distance = 180 km
Time = 3 h
🔄 Unit Conversion (for part ii)
km/h → m/s: multiply by 1000, divide by 3600.
Shortcut: divide km/h by 3.6 to get m/s.
📐 Formulas Used
(i, ii) Speed = Distance ÷ Time
(iii) Distance = Speed × Time
Step 1 — Speed in km/h
$$\text{Speed} = \frac{180 \text{ km}}{3 \text{ h}} = \boxed{60 \text{ km/h}}$$
Step 2 — Convert km/h to m/s
Multiply by 1000 (km → m) and divide by 3600 (h → s).
$$60 \text{ km/h} = \frac{60 \times 1000}{3600} = \boxed{\frac{50}{3} \approx 16.67 \text{ m/s}}$$
Step 3 — Distance in 4 h
$$\text{Distance} = 60 \text{ km/h} \times 4 \text{ h} = \boxed{240 \text{ km}}$$
Speed = 60 km/h = 16.67 m/s — distance in 4 h = 240 km.

Q 5
Question
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?
🔗 Concept:
8.3
Speed
Horse at 18 m/s vs train at 72 km/h — convert and compare.
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Train (20 m/s) is faster than the horse (18 m/s) by 2 m/s
Explanation
📋 Given
Horse speed = 18 m/s
Train speed = 72 km/h
🔄 Unit Conversion
Horse speed is in m/s. Train speed is in km/h.
Compare them in the same unit → convert train to m/s.
Divide km/h by 3.6 to get m/s.
📐 Conversion
m/s = km/h ÷ 3.6
Step 1 — Convert train speed to m/s
$$72 \text{ km/h} = \frac{72 \times 1000}{3600} = \boxed{20 \text{ m/s}}$$
Step 2 — Compare
Train = 20 m/s > Horse = 18 m/s.
$$\text{Difference} = 20 - 18 = \boxed{2 \text{ m/s}}$$
Train (20 m/s) > Horse (18 m/s) — train is faster by 2 m/s.

Q 6
Question
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.
🔗 Concept:
8.4
Uniform and Non-uniform Linear Motion
Uniform (highway) vs non-uniform (city traffic) — distinguish.
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Highway = uniform; city traffic = non-uniform
Explanation
Highway Car — Uniform Motion
Travels at constant speed.
Covers equal distances in equal time intervals.
City Car — Non-uniform Motion
Stops at signals, speeds up between them.
Covers unequal distances in equal time intervals.
Uniform = constant speed + equal distances per interval — non-uniform = varying speed + unequal distances.

Q 7
Question
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
40
50
60
70
Distance (m)
0
8
16
24
32
40
48
56
🔗 Concept:
8.4
Uniform and Non-uniform Linear Motion
Fill the gaps — use uniform motion rule (equal distance in equal time).
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Speed = 0.8 m/s — gaps filled: d(20s)=16 m, t(32m)=40 s, d(50s)=40 m, t(48m)=60 s
Explanation
Find speed: 8 m in 10 s → Speed = 0.8 m/s
Uniform motion: distance = 0.8 × time
$$d = 0.8 \times t$$
$$t = 20\text{ s} \Rightarrow d = 16\text{ m} \quad;\quad d = 32\text{ m} \Rightarrow t = 40\text{ s}$$
$$t = 50\text{ s} \Rightarrow d = 40\text{ m} \quad;\quad d = 48\text{ m} \Rightarrow t = 60\text{ s}$$
Speed = 0.8 m/s — use d = 0.8 × t to fill all gaps.

Q 8
Question
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.
🔗 Concepts:
8.4
Uniform and Non-uniform Linear Motion
|
8.3
Speed
Car: 60 km, 70 km, 50 km in 3 hours — is the motion uniform?
Show Answer
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Non-uniform motion — average speed = 60 km/h
Explanation
📋 Given
Hour 1: 60 km covered
Hour 2: 70 km covered
Hour 3: 50 km covered
Total time = 3 h
🔄 Unit Conversion
No conversion needed — all distances in km, time in h.
Answer will be in km/h.
📐 Formula
Average Speed = Total Distance ÷ Total Time
Step 1 — Check if motion is uniform
Distances in equal intervals: 60, 70, 50 km → all different.
Unequal distances →
Non-uniform motion
.
Step 2 — Find total distance
$$\text{Total distance} = 60 + 70 + 50 = 180 \text{ km}$$
Step 3 — Calculate average speed
$$\text{Average speed} = \frac{180 \text{ km}}{3 \text{ h}} = \boxed{60 \text{ km/h}}$$
Non-uniform (distances differ each hour) — average speed = 60 km/h.

Q 9
Question
Which type of motion is more common in daily life— uniform or non-uniform? Provide three examples from your experience to support your answer.
🔗 Concept:
8.4
Uniform and Non-uniform Linear Motion
Non-uniform motion is most common — provide three examples.
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Non-uniform motion is more common in daily life
Explanation
Uniform motion is an idealisation — rarely seen in practice.
Example 1:
A car in city traffic — stops at signals, speeds up between them.
Example 2:
A ball rolling down a slope — it speeds up continuously.
Example 3:
A person walking — pace changes with terrain and tiredness.
Non-uniform is more common — speed changes due to traffic, slopes, and other real-life factors.

Q 10
Question
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
🔗 Concepts:
8.4
Uniform and Non-uniform Linear Motion
|
8.3
Speed
Motion data over 100 s — uniform or non-uniform? Find average speed.
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Non-uniform motion — average speed = 0.6 m/s
Explanation
📋 Given
Total time = 100 s (measured every 10 s)
Total distance = 60 m (from data table)
Distances per 10 s: 6, 4, 6, 5, 8, 6, 7, 3, 10, 5 m
🔄 Unit Conversion
No conversion needed — time in s, distance in m.
Answer will be in m/s.
📐 Formula
Average Speed = Total Distance ÷ Total Time
Step 1 — Check if motion is uniform
Distances in 10 s intervals: 6, 4, 6, 5, 8, 6, 7, 3, 10, 5 m.
All different →
Non-uniform motion
.
Step 2 — Find average speed
$$\text{Average speed} = \frac{60 \text{ m}}{100 \text{ s}} = \boxed{0.6 \text{ m/s}}$$
Non-uniform (unequal distances each 10 s) — average speed = 0.6 m/s.

Q 11
Question
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?
🔗 Concepts:
8.3.1
Speed–Distance–Time
|
8.3
Speed
Vehicle: 2 km in segments — find speed for third segment and average speed.
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Speed of last segment = 20 m/s — average speed = 10 m/s
Explanation
📋 Given
Segment 1: distance = 500 m, speed = 10 m/s
Segment 2: distance = 500 m, speed = 5 m/s
Segment 3: distance = 1000 m, speed = v (unknown)
Total distance = 2000 m (= 2 km)
Total time = 200 s
🔄 Unit Conversion
No conversion needed — all distances in m, speeds in m/s.
📐 Formulas Used
Time = Distance ÷ Speed
Average Speed = Total Distance ÷ Total Time
Step 1 — Time for Segment 1
$$t_1 = \frac{500 \text{ m}}{10 \text{ m/s}} = 50 \text{ s}$$
Step 2 — Time for Segment 2
$$t_2 = \frac{500 \text{ m}}{5 \text{ m/s}} = 100 \text{ s}$$
Step 3 — Find time for Segment 3
Total time = 200 s. Time used = 50 + 100 = 150 s.
$$t_3 = 200 - 150 = 50 \text{ s}$$
Step 4 — Speed for last 1000 m
$$v = \frac{1000 \text{ m}}{50 \text{ s}} = \boxed{20 \text{ m/s}}$$
Step 5 — Average speed for whole journey
$$\text{Average speed} = \frac{2000 \text{ m}}{200 \text{ s}} = \boxed{10 \text{ m/s}}$$
Last segment v = 20 m/s — whole journey average speed = 10 m/s.

Construct a floating bowl-type water clock. Take a bowl, make a small hole at its bottom, and float it in a larger container of water. Water slowly enters through the hole until the bowl sinks. Experiment by using bowls of different sizes and making holes of different sizes, so that the time the bowl takes to sink can be close to 24 minutes.
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A floating bowl with a hole sinks after a fixed time, so it can measure a set interval.
Key Points
A bigger hole lets water in faster, so the bowl sinks sooner.
A larger bowl needs more water, so it takes longer to sink.
By adjusting the bowl size and hole size, the sinking time can be tuned close to 24 minutes.
Because the sinking time stays steady, it can be used to measure a fixed interval of time.
Ancient water clocks worked on exactly this idea.
A steady, repeatable sinking time turns the floating bowl into a simple water clock.

Design an activity to measure the pulse rate of your friends — the number of times the pulse beats in one minute. Then think of an activity in which you use your own pulse, instead of a clock, to measure time. Develop a short story around this idea — for example, a traveller with no watch who counts pulse beats to keep track of time.
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A resting pulse beats at a fairly steady rate, so it can act as a rough natural clock.
Key Points
Count the pulse beats for one minute using a stopwatch — this gives the pulse rate.
For most children the pulse rate is about 70 to 90 beats per minute.
At rest the pulse is fairly steady, so a fixed number of beats marks a fixed time.
A traveller with no watch could count pulse beats to estimate how much time has passed.
Long ago, scientists also used the pulse to time events before accurate clocks existed.
Because it repeats steadily, the pulse can be used as a simple measure of time.

In Activity 8.2 you measured the time period of a pendulum of a given length, and the different readings came out slightly different from each other. Find out what might cause these small differences. Think of ways to control those causes, then repeat the activity to check whether the difference between the readings becomes smaller.
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Small timing and counting errors cause the slight differences, and careful control reduces them.
Key Points
Starting or stopping the stopwatch a little early or late changes the reading.
Miscounting the number of oscillations also changes the result.
Air movement or pushing the bob instead of releasing it gently can affect the swing.
To control these: release the bob gently, count carefully, and time many oscillations together.
Timing 20 oscillations and dividing by 20 makes each reading more accurate.
When the activity is repeated carefully, the readings come much closer to each other.
Careful, repeated measurement reduces the small differences between pendulum readings.

Visit a playground that has a few swings. Measure the time taken by a swing to complete 10 oscillations and calculate its time period. Repeat this a few times with children of different weights to find out whether the time period stays almost the same. Then repeat with swings of different lengths, and find out how the time period changes as the length increases. Is a swing also an example of a pendulum?
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A swing behaves like a pendulum — its time period depends on length, not on weight.
Key Points
Measure the time for 10 oscillations and divide by 10 to get the time period.
With children of different weights, the time period stays almost the same.
With a longer swing, the time period becomes larger.
So the time period depends on the length of the swing, not on the weight of the child.
Yes — a swing is an example of a pendulum, because it oscillates about a mean position.
A swing is a pendulum: its time period grows with length and does not depend on weight.

Gather the timings of the winners of the 100 m, 200 m, and 400 m races, for both men and women, from the last two Olympic Games. Using the formula speed = distance ÷ time, calculate the speed of each winner. Compare the speeds across the three events, and find out in which event the runners reach the fastest speed.
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Calculating speed = distance ÷ time shows the sprint events have the highest average speeds.
Key Points
For each winner, divide the race distance by the winning time to get the speed.
The 100 m and 200 m sprints usually show the highest average speeds.
The 400 m race has a slightly lower average speed than the sprints.
This is because runners cannot keep their top speed for a longer distance.
Comparing the events shows how average speed depends on the race distance.
Speed = distance ÷ time shows that the shorter sprint races have the fastest average speeds.