-
Engineering and Architecture
Exams
Colleges
Predictors
Resources
-
Computer Application and IT
Quick Links
Colleges
-
Pharmacy
Colleges
Resources
-
Hospitality and Tourism
Colleges
Resources
Diploma Colleges
-
Competition
Other Exams
Resources
-
School
Exams
Ranking
Products & Resources
-
Study Abroad
Top Countries
Student Visas
-
Arts, Commerce & Sciences
Exams
Colleges
Upcoming Events
Resources
-
Management and Business Administration
Exams
Colleges & Courses
Predictors
-
Learn
Online Courses
Engineering Preparation
Medical Preparation
-
Online Courses and Certifications
Top Streams
Specializations
- Digital Marketing Certification Courses
- Cyber Security Certification Courses
- Artificial Intelligence Certification Courses
- Business Analytics Certification Courses
- Data Science Certification Courses
- Cloud Computing Certification Courses
- Machine Learning Certification Courses
- View All Certification Courses
Resources
-
Medicine and Allied Sciences
Colleges
Predictors
Resources
-
Law
Resources
Colleges
-
Animation and Design
Animation Courses
Colleges
Resources
-
Media, Mass Communication and Journalism
Exams
Colleges
Resources
-
Finance & Accounts
Top Courses & Careers
Colleges
Get Answers to all your Questions
Q 3.14 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 !]
Answers (1)
(a).(i) Time taken by the man to reach the market is t1
Displacement in 30 minutes is d1=2.5 km
Magnitude of average velocity=d1/t1= 5 km h-1
(b).(i) Distance travelled in 30 minutes is s1=2.5 km
Average speed=s1/t1= 5 km h-1
The first 30 minutes the man travels from his home to the market.
During the next 10 minutes, he travels with a speed of 7.5 km h-1 towards his home covering a distance s3
t3=40 min
Magnitude of average velocity =(2.5-1.25)/t3=1.875 km h-1
Average speed=(2.5+1.25)/t3=5.625 km h-1
Crack CUET with india's "Best Teachers"
- HD Video Lectures
- Unlimited Mock Tests
- Faculty Support
Similar Questions
- 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
- 1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.
- At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ?
Trending Articles/News
Latest Question
- A sum of money under compound interest doubles itself in 4 years. In how many years will it become 16 times itself? Option: 1 1
- A certain loan amounts, under compound interest, compounded annually earns an interest of Rs.1980 in the second year and Rs.2178 in the third year. How much interest did it earn in the first year?
