ICSE Class 8 Maths Commercial Arithmetic Chapter 13 Speed Distance Time

Speed, Distance, and Time

Topics covered:

Speed, Distance, and Time

Average Speed

Relative Speed

Speed, Distance, and Time

Let us recall the formulae used to calculate speed, distance, and time.

1. Speed = \[\frac{\text{Distance}}{\text{Time}}\]

If distance is constant, speed is inversely proportional to time, or if the distance to be covered is the same, the time taken will be inversely proportional to the speed. The more the speed, the less will be the time taken.

2. Distance = Speed × Time

If speed is constant, distance covered is directly proportional to the time taken, or if the same speed is maintained, the distance covered will be directly proportional to the time period. The more the time, the more will be the distance covered.

If time is constant, distance covered is directly proportional to the speed, or in the same time period, the distance covered will be directly proportional to the speed. The more the speed, the more will be the distance covered.

3. Time = \[\frac{\text{Distance}}{\text{Speed}}\]

If speed is constant, time is directly proportional to distance. The more the distance covered, the more will be the time taken. If distance is constant, time is inversely proportional to speed.

Teacher's Note

Understanding speed, distance, and time helps students calculate travel times for everyday activities like reaching school by bus, planning road trips, or understanding how fast different vehicles move.

Example 1

Express 9.9 km/h in m/s and 6.3 m/s in km/h.

1 km = 1000 m

or

\[1000 \text{ m} = 1 \text{ km} \Rightarrow 1 \text{ m} = \frac{1}{1000} \text{ km}\]

1 h = 60 minutes = 60 × 60 seconds = 3600 seconds

or

3600 s = 1 h

\[1 \text{ s} = \frac{1}{3600} \text{ h}\]

\[1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}\]

Conversely, \[1 \text{ m/s} = \frac{18}{5} \text{ km/h}\]

\[9.9 \text{ km/h} = 9.9 \times \frac{5}{18} = 2.75 \text{ m/s}\]

\[6.3 \text{ m/s} = 6.3 \times \frac{18}{5} \text{ km/h}\]

= 22.68 km/h

To convert from km/h to m/s, multiply by \[\frac{1000}{3600}\] or \[\frac{5}{18}\]

To convert from m/s to km/h, multiply by \[\frac{3600}{1000}\] or \[\frac{18}{5}\]

Example 2

A night bus needs to cover 408 km in 8 h. If it covers the first quarter of the journey at 45 km/h and the second quarter of the journey at 55 km/h, what should be its speed for the rest of the journey?

Distance covered in the first quarter = 45 km/h × 2 h = 90 km

Distance covered in the second quarter = 55 km/h × 2 h = 110 km

Distance to be covered in 4 h = 408 - (90 + 110) = 208 km

Speed for the remaining journey = \[\frac{\text{Distance remaining}}{\text{Time remaining}} = \frac{208 \text{ km}}{4 \text{ h}} = 52 \text{ km/h}\]

Example 3

The Coalfield Express, 150 m long, was waiting at Durgapur station when the New Delhi-bound Rajdhani Express, 130 m long, thundered past it in only 8 seconds. At what speed was the Rajdhani Express travelling?

In order to go past the Coalfield Express the engine of Rajdhani Express has to go past its entire length and then go another 130 m till its own last compartment clears the Coalfield Express.

Thus distance travelled = 150 m + 130 m = 280 m

Time taken = 8 seconds

Speed = \[\frac{280}{8} = 35 \text{ m/s} = 35 \times \frac{18}{5} = 126 \text{ km/h}\]

Thus, the Rajdhani Express thundered past the Coalfield Express at 126 km/h.

Example 4

Two racing cars are flagged off together. Car A travels at 124.2 km/h, while car B travels at 120 km/h to finish the race 7 seconds after car A. What was the distance covered in the race?

Method I: Using Direct Proportion

\[124.2 \text{ km/h} = 124.2 \times \frac{5}{18} = 34\frac{1}{2} \text{ m/s}\]

\[120 \text{ km/h} = 120 \times \frac{5}{18} = 33\frac{1}{3} \text{ m/s}\]

Speed = \[\frac{\text{Distance}}{\text{Time}} \Rightarrow \text{Time} = \frac{\text{Distance}}{\text{Speed}}\]

At \[34\frac{1}{2}\] m/s, covering 1000 m would take

\[1000 \div 34\frac{1}{2} = \frac{2000}{69}\] seconds

At \[33\frac{1}{3}\] m/s, covering 1000 m would take

\[1000 \div 33\frac{1}{3} = 30\] seconds

Thus, difference in time taken to cover 1000 m

= \[30 - \frac{2000}{69} = \frac{70}{69}\] seconds

Actual difference in time taken to cover x m = 7 seconds

Direct proportion = Time difference for 1000 m : Time difference for x m : : 1000 m : x m

\[\frac{70}{69} : 7 : : 1000 : x\]

\[\Rightarrow \frac{70x}{69} = 7 \times 1000\]

\[\Rightarrow x = \frac{7000 \times 69}{70} = 6900 \text{ m} = 6.9 \text{ km}\]

Method II: Using Formula

Let the distance covered in the race be represented by x.

Covering x m at \[34\frac{1}{2}\] m/s = \[x \div \frac{69}{2} = \frac{2x}{69}\] seconds

Covering x m at \[33\frac{1}{3}\] m/s = \[x \div \frac{100}{3} = \frac{2x}{100}\] seconds

Time difference = \[\frac{3x}{100} - \frac{2x}{69} - \frac{7x}{6900}\] seconds

Given that \[\frac{7x}{6900} = 7\] seconds

\[\Rightarrow x = \frac{7 \times 6900}{7} = 6900 \text{ m} = 6.9 \text{ km}\]

Teacher's Note

Racing problems help students understand how small differences in speed lead to differences in race completion times, which is relevant to sports and competitive activities.

Average Speed

Most moving bodies that we see around us rarely run at a fixed speed. Sometimes they slow down and at other times they travel faster. In such instances we need to find the average speed.

Average speed = \[\frac{\text{Total distance covered}}{\text{Total time taken}}\]

Example 5

Arijit started off eagerly and ran the first 350 m of the race at 25.2 km/h. Then settling into a steady 18 km/h, he ran the next 1000 m. He made a dash for it towards the finishing line, running the last 150 m at 27 km/h. How much time did he take to finish the race and what was his average speed?

\[25.2 \text{ km/h} = 25.2 \times \frac{5}{18} = 7 \text{ m/s}\]

First 350 m at 25.2 km/h or 7 m/s took \[\frac{350}{7} = 50\] seconds

Next 1000 m at 18 km/h or 5 m/s took \[\frac{1000}{5} = 200\] seconds

Last 150 m at 27 km/h or 7.5 m/s took \[\frac{150}{7.5} = 20\] seconds

Thus covering 1500 m took the runner = 270 seconds

Total distance covered

Average speed = \[\frac{\text{Total distance covered}}{\text{Total time taken}}\]

= \[\frac{1500 \text{ m}}{270 \text{ s}} = 5\frac{5}{9} \text{ m/s}\]

= \[\frac{50}{9} \text{ m/s} = \frac{50}{9} \times \frac{18}{5} = 20 \text{ km/h}\]

Teacher's Note

Average speed calculations are useful for understanding how runners maintain different paces during a race or how drivers adjust speeds based on road conditions.

Relative Speed

1. If you float on water in a swimming pool, you will stay in the same place because the water in the swimming pool is not going anywhere.

2. But if you float on water in a river, you will be carried downstream at the speed of the river's current. Although your speed will be 0, your speed in relation to the river bank, or your relative speed, will be the speed of the river's current.

3. If you swim downstream, your speed in relation to the river bank will be more than the speed at which you are swimming, as your relative speed will be your speed plus the speed of the river's current.

4. If you swim upstream against the river's current at a speed equal to that of the river's current, your position in relation to the river bank will not change as your relative speed will be 0. This is because the water's opposite speed will cancel out your effort.

5. If you swim upstream against the river's current at a speed greater than the river's current, you will surely make some headway as your relative speed will be your speed minus the speed of the river's current.

Example 6

A bird is flying at 57.6 km/h against a strong wind blowing at 7.2 km/h. How long would it take to fly from one tree top to another 630 m away?

Speed of bird = \[57.6 \times \frac{5}{18} = 16 \text{ m/s}\]

Opposite speed of wind = \[7.2 \times \frac{5}{18} = 2 \text{ m/s}\]

Relative speed of bird = 16 m/s - 2 m/s = 14 m/s

Time taken = \[\frac{\text{Distance to be covered}}{\text{Relative speed}} = \frac{630}{14} = 45 \text{ s}\]

Example 7

The engine driver of a 130 m long Rajdhani Express, travelling at 144 km/h, sees the rear of a slow-moving goods train ahead of him on another track. If the Rajdhani Express takes 19 seconds to overtake the goods train which was travelling at 75.6 km/h, how long was the goods train?

Speed of overtaking = Relative speed = Speed of Rajdhani Express - Speed of goods train

= 144 km/h - 75.6 km/h = 68.4 km/h

= \[68.4 \times \frac{5}{18} = 19 \text{ m/s}\]

Distance covered = Speed × Time = 19 m/s × 19 s = 361 m

Length of goods train = 361 m - Length of Rajdhani Express = 361 m - 130 m = 231 m

Teacher's Note

Relative speed concepts are essential for understanding overtaking in traffic, river crossing problems, and how wind affects the movement of flying objects.

Try this!

1. A scooterist covers a certain distance at 36 kmph. How many metres does he cover in 2 minutes?

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