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  • A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h –1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the

    a) magnitude of average velocity, and

    b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?

    [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]

A man walks on a straight road from his | Class 11 Physics Chapter Motion in a straight Line, Motion in a straight Line NCERT Solutions

Q14.

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h –1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the

a) magnitude of average velocity, and

b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?

[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]

(i)

Time taken by the man to reach the market from home,  t1  =  2.5/5  =  ½ h = 30 min

Time taken by the man to reach home from the market, t2  = 2.5/7.5  =  1/3 h  =  20 min

Total time taken in the whole journey = 30 + 20 = 50 min

Average velocity  =  Displacement / Time  =  2.5 / ½  =  5 km/h                ..... (a(i))

Average speed   =  Distance /  time  =  2.5 / ½  =  5 km/h                ..... (b(i))

 

(ii) Time = 50 min =  5/6 h

Net displacement = 0

Total distance = 2.5 + 2.5 = 5 km

Average velocity  =  Displacement / Time  =  0             ..... (a(ii))

Average speed   =  Distance /  time  = 5 / 5/6  =  6 km/h           ..... (b(ii))

 

(iii)

Speed of the man = 7.5 km

Distance travelled in first 30 min = 2.5 km

Distance travelled by the man (from market to home) in the next 10 min 

= 7.5 x 10/60 = 1.25 km

Net displacement = 2.5 – 1.25 = 1.25 km

Total distance travelled = 2.5 + 1.25 = 3.75 km

Average Velocity  =  1.25 / 40/60  =  1.25x3 / 2  =  1.875km/h   ...........  (a(iii))

Average Speed =  3.75 / 40/60 = 5.625 km/h     ..... (b(iii))

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Important Questions & Answers

Why is this answer important for exams?

This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.

Which NCERT concept is used in this question?

This question is based on core NCERT concepts explained in the chapter and should be revised thoroughly before exams.

What common mistakes should be avoided in this question?

Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

What is the correct answer to: A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h –1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the a) magnitude of average velocity, and b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]?

(i)

Time taken by the man to reach the market from home,  t1  =  2.5/5  =  ½ h = 30 min

Time taken by the man to reach home from the market, t2  = 2.5/7.5  =  1/3 h&nb...

How do you solve A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h –1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the a) magnitude of average velocity, and b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!] step by step?

Step-by-step explanation:
• (i)

• Time taken by the man to reach the market from home,  t1  =  2
• 5/5  =  ½ h = 30 min

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