ICSE Class 8 Maths Chapter 06 Speed Time and Distance

Chapter 6

Speed, Time And Distance

You have already solved simple problems on speed, distance and time in the previous class. In this chapter, we shall develop those ideas a little further. We shall also take up problems on relative speed and average speed.

Speed of an object is the distance covered by it in a unit time.

For Example:

(i) If a car covers a distance of 195 km in 3 hours, its speed is \(\frac{195 \text{ km}}{3 \text{ hour}}\) i.e. 65 km per hour. It is written as 65 km/hr.

(ii) If an aeroplane covers an aerial distance of 1750 km in \(2\frac{1}{2}\) hours, its speed = \(\frac{1750}{5/2}\) km/hr = \(\left(1750 \times \frac{2}{5}\right)\) km/hr = 700 km/hr.

(iii) If an athelete covers a distance of 750 metres in 1 minute 40 seconds i.e. in 100 seconds, then his speed = \(\frac{750}{100}\) metres per second = 7.5 m/sec.

Thus,

\[\text{speed} = \frac{\text{distance}}{\text{time}}\]

From this formula, we get

\[\text{distance} = \text{speed} \times \text{time}\]

\[\text{time} = \frac{\text{distance}}{\text{speed}}\]

Speed is usually given in km/hr (kilometres per hour) or m/sec (metres per second). Note that unit of measuring must be mentioned.

Conversion Of Units Of Speed

To convert km/hr to m/sec, note that

\[1 \text{ km/hr} = \frac{1 \text{ km}}{1 \text{ hour}} = \frac{1000 \text{ metres}}{(60 \times 60) \text{ seconds}} = \frac{5}{18} \text{ m/sec.}\]

Hence, to convert km/hr into m/sec, multiply by \(\frac{5}{18}\).

To convert m/sec to km/hr, note that

\[1 \text{ m/sec} = \frac{1 \text{ m}}{1 \text{ sec}} = \frac{\frac{1}{1000} \text{ km}}{\frac{1}{60 \times 60} \text{ hour}} = \frac{3600}{1000} \text{ km/hr} = \frac{18}{5} \text{ km/hr.}\]

Hence, to convert m/sec into km/hr, multiply by \(\frac{18}{5}\).

Teacher's Note

Speed calculations are used daily when driving, flying, or commuting - understanding unit conversions helps us interpret real-world speed limits and travel times accurately.

Uniform Speed, Variable Speed

If an object covers equal distances in equal intervals of time, its speed is said to be uniform (or constant); otherwise, its speed is said to be variable.

The above formulae connecting speed, distance and time are based on the assumption that the speed is uniform.

Average Speed

Most of the time, a vehicle does not cover the entire distance at a uniform speed. It picks up speed, then covers some distance and then reduces speed to come to halt. In such a case, we calculate the total distance travelled and divide it by the total time taken to get average speed.

\[\text{Average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}\]

Remark

To pass a fixed object (post, pole or man etc.), a train has to cover a distance = length of the train.

To pass a bridge (tunnel or platform), a train has to cover a distance = length of the train + length of bridge (tunnel or platform).

If one train passes another train (in the same direction or in the opposite direction), the distance covered = sum of lengths of both the trains.

Example 1.

A cyclist covers 600 metres in 5 minutes.

(i) Calculate his speed in m/sec as well as in km/hr.

(ii) How much distance will he travel in \(1\frac{1}{2}\) hours?

(iii) How much time will he require to cover 1.8 km?

Solution.

(i) Distance covered = 600 metres, time taken = 5 minutes = (5 × 60) seconds = 300 seconds

\(\therefore\) speed = \(\frac{\text{distance}}{\text{speed}}\) = \(\frac{600 \text{ m}}{300 \text{ sec}}\) = 2 m/sec

In km/hr, speed = \(\left(2 \times \frac{18}{5}\right)\) km/hr = 7.2 km/hr.

(ii) Distance covered in \(1\frac{1}{2}\) hours = speed × time = (7.2 km/hr) × 1.5 hours = 10.8 km.

(iii) Time required to cover 1.8 km = \(\frac{\text{distance}}{\text{speed}}\) = \(\frac{1.8 \text{ km}}{7.2 \text{ km/hr}}\) = \(\frac{18}{72}\) hr = \(\frac{1}{4}\) hr = 15 minutes.

Example 2.

A car completes a journey of 360 km in \(5\frac{3}{4}\) hours. If it covers the first three-fourth of the journey at 60 km/hr, find the speed of the car for the rest of the journey.

Solution.

Total distance of the journey = 360 km

Distance of three-fourth of the journey = \(\left(\frac{3}{4} \times 360\right)\) km = 270 km

Time taken to cover this part of journey = \(\frac{\text{distance}}{\text{speed}}\) = \(\frac{270 \text{ km}}{60 \text{ km/hr}}\) = \(\frac{9}{2}\) hours = \(4\frac{1}{2}\) hours

Remaining distance of the journey = (360 - 270) km = 90 km

Remaining time = \(\left(5\frac{3}{4} - 4\frac{1}{2}\right)\) hours = \(\left(\frac{23}{4} - \frac{9}{2}\right)\) hours = \(\frac{23 - 18}{4}\) hours = \(\frac{5}{4}\) hours

\(\therefore\) Speed during the rest of journey = \(\frac{\text{distance}}{\text{speed}}\) = \(\frac{90 \text{ km}}{\frac{5}{4} \text{ hours}}\) = \(\left(90 \times \frac{4}{5}\right)\) km/hr = 72 km/hr.

Teacher's Note

Journey problems teach us to break complex trips into segments, calculating speeds for different portions - a skill useful when planning real trips with varying road conditions.

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