Measurement of Time and Motion

Given:
- Distance = 150 m
- Time = 10 s

Formula:
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$

Step 1: Find speed in m/s
$$\text{Speed} = \frac{150 \text{ m}}{10 \text{ s}} = 15 \text{ m/s}$$

Step 2: Convert m/s to km/h
$$1 \text{ m/s} = \frac{3600}{1000} \text{ km/h} = 3.6 \text{ km/h}$$
$$\text{Speed} = 15 \times 3.6 = 54 \text{ km/h}$$

Answer: The speed of the car is $\mathbf{54 \text{ km/h}}$.

Given:
- Runner 1: Distance = 400 m, Time = 50 s
- Runner 2: Distance = 400 m, Time = 45 s

Formula:
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$

Speed of Runner 1:
$$v_1 = \frac{400}{50} = 8 \text{ m/s}$$

Speed of Runner 2:
$$v_2 = \frac{400}{45} = \frac{80}{9} \approx 8.89 \text{ m/s}$$

Comparison:
Since v_2 > v_1, Runner 2 is faster.

Difference in speed:
$$v_2 - v_1 = \frac{80}{9} - 8 = \frac{80 - 72}{9} = \frac{8}{9} \approx 0.89 \text{ m/s}$$

Answer: Runner 2 has a greater speed. Runner 2 is faster than Runner 1 by approximately $\mathbf{\frac{8}{9} \approx 0.89 \text{ m/s}}$.

Given:
- Speed = 25 m/s
- Distance = 360 km

Step 1: Convert distance to metres
$$360 \text{ km} = 360 \times 1000 = 360000 \text{ m}$$

Formula:
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

Step 2: Calculate time
$$\text{Time} = \frac{360000 \text{ m}}{25 \text{ m/s}} = 14400 \text{ s}$$

Step 3: Convert to hours
$$14400 \text{ s} = \frac{14400}{3600} \text{ h} = 4 \text{ h}$$

Answer: The train takes $\mathbf{14400 \text{ s}}$ (i.e., 4 hours) to cover 360 km.

Given:
- Distance = 180 km
- Time = 3 h

(i) Speed in km/h:
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{180 \text{ km}}{3 \text{ h}} = 60 \text{ km/h}$$

(ii) Speed in m/s:
$$60 \text{ km/h} = \frac{60 \times 1000}{3600} \text{ m/s} = \frac{60000}{3600} = \frac{50}{3} \approx 16.67 \text{ m/s}$$

(iii) Distance in 4 h at the same speed:
$$\text{Distance} = \text{Speed} \times \text{Time} = 60 \text{ km/h} \times 4 \text{ h} = 240 \text{ km}$$

Answers:
- (i) Speed = $\mathbf{60 \text{ km/h}}$
- (ii) Speed = $\mathbf{\frac{50}{3} \approx 16.67 \text{ m/s}}$
- (iii) Distance in 4 h = $\mathbf{240 \text{ km}}$

Given:
- Speed of horse = 18 m/s
- Speed of train = 72 km/h

Step 1: Convert train speed to m/s
$$72 \text{ km/h} = \frac{72 \times 1000}{3600} \text{ m/s} = \frac{72000}{3600} = 20 \text{ m/s}$$

Step 2: Compare
- Speed of horse = 18 m/s
- Speed of train = 20 m/s

$$\text{Difference} = 20 - 18 = 2 \text{ m/s}$$

Answer: The train is faster than the fastest galloping horse. The train moves at 20 m/s while the horse moves at 18 m/s, so the train is faster by $\mathbf{2 \text{ m/s}}$.

Uniform Motion:
When an object covers equal distances in equal intervals of time, it is said to be in uniform motion.

*Example:* A car moving on a straight highway with no traffic maintains a constant speed (say 80 km/h). It covers equal distances in equal time intervals. This is uniform linear motion.

Non-Uniform Motion:
When an object covers unequal distances in equal intervals of time, it is said to be in non-uniform motion.

*Example:* A car moving in city traffic has to slow down at signals, speed up on clear stretches, and stop at crossings. It covers different distances in equal time intervals. This is non-uniform motion.

| Feature | Uniform Motion | Non-Uniform Motion |
|---|---|---|
| Speed | Constant | Keeps changing |
| Distance in equal time | Equal | Unequal |
| Example | Car on empty highway | Car in city traffic |

Given: The object is in uniform motion, so it covers equal distances in equal time intervals.

Step 1: Find the speed (rate of distance covered)
From the table: In 10 s, distance covered = 8 m
$$\text{Speed} = \frac{8 \text{ m}}{10 \text{ s}} = 0.8 \text{ m/s}$$

Step 2: Verify with known values
- At t = 30 s: Distance = $0.8 \times 30 = 24$ m ✓
- At t = 50 s: Distance = $0.8 \times 50 = 40$ m ✓
- At t = 70 s: Distance = $0.8 \times 70 = 56$ m ✓

Step 3: Fill in the gaps
- At t = 20 s: Distance = $0.8 \times 20 = \mathbf{16 \text{ m}}$
- Missing time when Distance = 32 m: $t = \frac{32}{0.8} = \mathbf{40 \text{ s}}$
- At t = 60 s: Distance = $0.8 \times 60 = \mathbf{48 \text{ m}}$

Completed Table:

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

Given:
- Distance in 1st hour = 60 km
- Distance in 2nd hour = 70 km
- Distance in 3rd hour = 50 km

Is the motion uniform?
In uniform motion, equal distances must be covered in equal time intervals. Here, the distances covered in each hour are different (60 km, 70 km, 50 km). Since the distances are unequal in equal time intervals (1 hour each), the motion is non-uniform.

Average Speed:
$$\text{Total distance} = 60 + 70 + 50 = 180 \text{ km}$$
$$\text{Total time} = 1 + 1 + 1 = 3 \text{ h}$$
$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{180 \text{ km}}{3 \text{ h}} = 60 \text{ km/h}$$

Answer: The motion is non-uniform. The average speed of the car is $\mathbf{60 \text{ km/h}}$.

Answer: Non-uniform motion is more common in daily life.

In most real-life situations, objects change their speed due to traffic, obstacles, turns, or varying forces. Perfectly constant speed over a long duration is rare.

Three examples of non-uniform motion from daily life:

1. A bus on a city road: A bus speeds up after a bus stop, slows down at traffic signals, and stops at the next bus stop. It covers unequal distances in equal time intervals.

2. A ball rolling on the ground: When a ball is rolled on the ground, it gradually slows down due to friction and eventually stops. Its speed keeps decreasing, so it covers unequal distances in equal time intervals.

3. A person walking: When a person walks, they may walk faster on an open path and slower in a crowded area or while climbing stairs. Their speed keeps changing, making the motion non-uniform.

Conclusion: Non-uniform motion is far more common in daily life because most moving objects experience changing forces (friction, air resistance, traffic, etc.) that cause their speed to vary.

Step 1: Check distances covered in each 10-second interval

| Interval (s) | Distance covered (m) |
|---|---|
| 0 to 10 | 6 − 0 = 6 |
| 10 to 20 | 10 − 6 = 4 |
| 20 to 30 | 16 − 10 = 6 |
| 30 to 40 | 21 − 16 = 5 |
| 40 to 50 | 29 − 21 = 8 |
| 50 to 60 | 35 − 29 = 6 |
| 60 to 70 | 42 − 35 = 7 |
| 70 to 80 | 45 − 42 = 3 |
| 80 to 90 | 55 − 45 = 10 |
| 90 to 100 | 60 − 55 = 5 |

The distances covered in equal time intervals (10 s each) are not equal. Therefore, the motion is non-uniform.

Step 2: Find average speed
$$\text{Total distance} = 60 \text{ m}$$
$$\text{Total time} = 100 \text{ s}$$
$$\text{Average speed} = \frac{60 \text{ m}}{100 \text{ s}} = 0.6 \text{ m/s}$$

Answer: The speed of the object is non-uniform. The average speed is $\mathbf{0.6 \text{ m/s}}$.

Given:
- Total distance = 2 km = 2000 m
- First 500 m at speed = 10 m/s
- Next 500 m at speed = 5 m/s
- Total time for journey = 200 s

Step 1: Find time taken for first 500 m
$$t_1 = \frac{500}{10} = 50 \text{ s}$$

Step 2: Find time taken for next 500 m
$$t_2 = \frac{500}{5} = 100 \text{ s}$$

Step 3: Find remaining time for the rest of the journey
$$t_3 = 200 - t_1 - t_2 = 200 - 50 - 100 = 50 \text{ s}$$

Step 4: Find remaining distance
$$\text{Remaining distance} = 2000 - 500 - 500 = 1000 \text{ m}$$

Step 5: Find required speed for remaining distance
$$v_3 = \frac{1000 \text{ m}}{50 \text{ s}} = 20 \text{ m/s}$$

Step 6: Find average speed for entire journey
$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2000 \text{ m}}{200 \text{ s}} = 10 \text{ m/s}$$

Answers:
- The vehicle must cover the remaining 1000 m at a speed of $\mathbf{20 \text{ m/s}}$.
- The average speed for the entire journey is $\mathbf{10 \text{ m/s}}$.