ICSE Class 7 Maths Chapter 07 Speed Distance Time

Chapter 7

Speed, Distance and Time

7.1 Speed

Speed of a body is the distance covered by the body in unit time.

Speed = \(\frac{\text{Distance}}{\text{Time}}\)

(i) Distance = Speed × Time

(ii) Time = \(\frac{\text{Distance}}{\text{Speed}}\)

1. In order to find speed, if:

(i) distance is in metre (m) and time in second (s); then the speed is in metre per second (m s-1).

(ii) distance is in kilometre (km) and time in hour (h); then the speed is in kilometre per hour (km h-1).

2. In order to find distance, if:

(i) speed is in m s-1, time must be in second.

(ii) speed is in km h-1, time must be in hour.

3. In order to find time, if:

(i) speed is in km h-1, distance must be in kilometre.

(ii) speed is in m s-1, distance must be in metre.

Example 1

A boy covers a distance of 1.2 km in 40 minutes. Find his speed in:

(i) km per hour (km h-1)

(ii) metre per second (m s-1)

Solution

(i) In order to get speed in km per hour; the distance covered must be in km and the time taken must be in hour.

Given: distance = 1.2 km and time = 40 min = \(\frac{40}{60}\) h = \(\frac{2}{3}\) h

Speed = \(\frac{\text{Distance}}{\text{Time}}\)

= \(\frac{1.2 \text{ km}}{\frac{2}{3} \text{ h}}\) = 1.2 × \(\frac{3}{2}\) km h-1 = 1.8 km h-1

(ii) In order to get speed in metre per second; the distance covered must be in metre and the time taken must be in second.

Given: distance = 1.2 km = 1.2 × 1000 m = 1,200 m

And, time = 40 min = 40 × 60 sec = 2400 sec

Speed = \(\frac{\text{Distance}}{\text{Time}}\)

= \(\frac{1200 \text{ m}}{2400 \text{ sec}}\) = \(\frac{1}{2}\) m s-1 = 0.5 m s-1

Teacher's Note

When you travel to school by bus, the bus's speed determines whether you arrive on time or late; understanding speed helps us plan our journeys better.

7.2 Uniform Speed and Variable Speed

If a body covers equal distances in equal intervals of time, its speed is said to be uniform otherwise its speed is variable.

For example

(i) If a car covers 60 km in first hour, 60 km in second hour, 60 km in third hour and so on, its speed is uniform.

(ii) If a car covers 60 km in first hour, 67 km in second hour, 58 km in third hour and so on, its speed is variable.

(iii) If a car cover first 60 km in one hour, second 60 km in 1 hour 20 minutes, third 60 km in 1 hour 30 minutes and so on, then also its speed is variable.

Example 2

A man runs 200 metre in 25 second. Find:

(i) his speed

(ii) the distance run by him in 5 seconds

(iii) the time taken by him to cover \(\frac{2}{5}\) km.

Solution

(i) Speed = \(\frac{\text{Distance}}{\text{Time}}\) = \(\frac{200 \text{ m}}{25 \text{ sec}}\) = 8 m s-1

(ii) Distance run in 5 sec = Speed × Time = 8 m s-1 × 5 sec = 40 m

(iii) Time taken to cover \(\frac{2}{5}\) km = \(\frac{\text{Distance}}{\text{Speed}}\)

= \(\frac{400 \text{ m}}{8 \text{ m s}^{-1}}\)

[Note: \(\frac{2}{5}\) km = \(\frac{2}{5}\) × 1000 m = 400 m]

= 50 seconds

Teacher's Note

When athletes train for races, they monitor their speed and distance covered in practice sessions to improve their performance for competitions.

Example 3

A train covers first 120 km in 2 hours, next 160 km in 3 hours and last 140 km again in 2 hours. Find the average speed of the train.

Solution

Average speed of an object = \(\frac{\text{Total distance covered by it}}{\text{Total time taken}}\)

Since, total distance covered = 120 km + 160 km + 140 km = 420 km

And, total time taken = 2 hr + 3 hr + 2 hr = 7 hr.

Average speed = \(\frac{420 \text{ km}}{7 \text{ hr}}\) = 60 km h-1

Example 4

A man covers first 60 km of his journey at 30 km h-1 and remaining 50 km at 20 km h-1. Find: (i) the total time taken, (ii) his average speed during the whole journey.

Solution

(i) Time taken to cover 1st 60 km = \(\frac{60}{30}\) h = 2 h

[Note: Time = \(\frac{\text{Distance}}{\text{Speed}}\)]

And, time taken to cover remaining 50 km = \(\frac{50}{20}\) h = \(\frac{5}{2}\) h

Total time taken = 2 h + \(\frac{5}{2}\) h = \(\frac{9}{2}\) h = 4\(\frac{1}{2}\) h

(ii) Since, total distance covered = 60 km + 50 km = 110 km

and total time taken = \(\frac{9}{2}\) h

Average speed = \(\frac{110}{\frac{9}{2}}\) km h-1

[Note: Average speed = \(\frac{\text{Total distance covered}}{\text{Total time taken}}\)]

= \(\frac{110 × 2}{9}\) km h-1 = 24\(\frac{4}{9}\) km h-1

Teacher's Note

When planning a road trip, calculating average speed helps us estimate arrival time even though we may travel at different speeds through various sections of the journey.

Exercise 7 (A)

1. Fill in the blanks:

(i) A distance of 40 m is covered in 8 sec => speed = .............. m/s.

(ii) A distance of 1.4 km is covered in 10 min => speed = .............. km/min.

(iii) A distance of 32 km is covered in 1.6 hrs => speed = .............. km/h.

(iv) A car moves at 60 km h-1 for 40 min => distance covered = .............. km.

(v) If speed = 30 m s-1, distance covered in 15 min = .............. m = .............. km.

(vi) Speed = 15 km min-1 and time = 1 hour => distance covered = .............. km.

(vii) If speed = 1.2 km min-1 and distance covered = 36 km; time taken ............ min.

(viii) If speed = 18 m sec-1 and distance covered = 2.7 km; time taken = ...................... sec = ..................... min.

2. A train covers 51 km in 3 hours. Calculate its speed. How far does the train go in 30 minutes ?

3. A motorist travelled the distance between two towns, which is 65 km in 2 hours and 10 minutes. Find his speed in metre per minute.

4. A train travels 700 metres in 35 seconds. What is its speed in km h-1 ?

5. A racing car covered 600 km in 3 hours 20 minutes. Find its speed in metre per second. How much distance will the car cover in 50 sec ?

6. Rohit goes 350 km in 5 hours. Find:

(i) his speed

(ii) the distance covered by Rohit in 6.2 hours

(iii) the time taken by him to cover 210 km, [Assume that throughout the journey, the speed of Rohit remains uniform].

7. A boy drives his scooter with a uniform speed of 45 km h-1. Find:

(i) the distance covered by him in 1 hour 20 min.

(ii) the time taken by him to cover 108 km.

(iii) the time taken to cover 900 m.

8. I travel a distance of 10 km and come back in 2 \(\frac{1}{2}\) hours. What is my speed ?

9. A man walks a distance of 5 km in 2 hours. Then he goes in a bus to a nearby town, which is 40 km in further 2 hours. From there, he goes to his office in an autorickshaw, a distance of 5 km in \(\frac{1}{2}\) hour. What was his average speed during the whole journey.

10. Jagan went to another town such that he covered 240 km by a car going at 60 km h-1. Then he covered 80 km by a train, going at 100 km h-1 and the rest 200 km, he covered by a bus, going at 50 km h-1. What was his average speed during the whole journey ?

11. The speed of sound in air is about 330 m s-1. Express this speed in km h-1. How long will the sound take to travel 99 km ?

7.3 Converting Speed From One Unit to Other Unit

To convert speed in kilometre per hour (km h-1) into metre per second (m s-1), multiply by \(\frac{5}{18}\). And, to convert m s-1 into km h-1, multiply by \(\frac{18}{5}\).

Reason: 1 km h-1 = \(\frac{1 \text{ kilometre}}{1 \text{ hour}}\) = \(\frac{1000 \text{ metre}}{60 × 60 \text{ second}}\) = \(\frac{5}{18}\) m s-1.

Example 5

Convert: (i) 90 km h-1 into m s-1 (ii) 15 m s-1 into km h-1 (iii) 75 cm s-1 into km h-1 (iv) 45 km h-1 into m min-1

Solution

(i) 90 km h-1 = 90 × \(\frac{5}{18}\) ms-1 = 25 m s-1

(ii) 15 m s-1 = 15 × \(\frac{18}{5}\) km h-1 = 54 km h-1

(iii) 75 cm s-1 = 0.75 m s-1

[Note: 75 cm = \(\frac{75}{100}\) m = 0.75 m]

= 0.75 × \(\frac{18}{5}\) km h-1 = 2.7 km h-1

(iv) 45 km h-1 = \(\frac{45 \text{ km}}{1 \text{ h}}\) = \(\frac{45 × 1000 \text{ m}}{60 \text{ min}}\) = 750 m min-1

When a train passes a:

(i) pole or any other stationary object, etc., the minimum distance covered by the train = length of the train

(ii) platform, the minimum distance covered by the train = length of the train + length of the platform.

Example 6

A 160 m long train is travelling at a speed of 72 km h-1, find the time taken by the train to pass:

(i) a telegraph post

(ii) a 200 m long platform.

Solution

(i) Distance to be covered = length of the train = 160 m

And, speed = 72 km h-1 = 72 × \(\frac{5}{18}\) m s-1 = 20 m s-1

Time taken = \(\frac{\text{Distance}}{\text{Speed}}\) = \(\frac{160}{20}\) sec = 8 sec

(ii) Distance to be covered = length of the train + length of the platform = 160 m + 200 m = 360 m

Time taken = \(\frac{\text{Distance}}{\text{Speed}}\) = \(\frac{360}{20}\) sec = 18 sec

Example 7

P and Q run with speeds 8 km h-1 and 11 km h-1. They start running from the same point, find the distance between them after 2 hours, if they run in the (i) same direction. (ii) opposite directions (moving away from each other).

Solution

(i) Required distance = Difference between the distances covered by P and Q

(ii) Required distance = Sum of the distances covered by P and Q.

Distance run by P in 2 hours = speed × time = 8 km h-1 × 2 hours = 16 km

Distance run by Q in 2 hours = 11 km h-1 × 2 hours = 22 km

(i) Required distance = 22 km - 16 km = 6 km

(ii) Required distance = 22 km + 16 km = 38 km

Teacher's Note

Understanding relative speed helps explain why two cars on the highway appear to move closer or farther apart depending on whether they travel in the same or opposite directions.

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