A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed he instantly turns and walks back home with a speed of 7.5 km/h. Calculate his average speed at the end of 50 minutes.
Share
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed he instantly turns and walks back home with a speed of 7.5 km/h. Calculate his average speed at the end of 50 minutes.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
Time taken by the man to reach the market from home,t1 = 2.5/5 = 1/2 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
(i) 0 to 30 min
Average velocity = Displacement/Time = 2.5/(1/2) = 5 km/h
Average speed = Distance/Time = 2.5/(1/2) = 5 km/h
(ii) 0 to 50 min
Time = 50 min = 50/60 = 5/6 h
Net displacement = 0
Total distance = 2.5 + 2.5 = 5 km
Average velocity = Displacement / Time = 0
Average speed = Distance / Time = 5/(5/6) = 6 km/h
(iii) 0 to 40 min
Speed of the man = 7.5 km/h
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 × 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 = Displacement / Time = 1.25 / (40/60) = 1.875 km/h
Average speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h
Explanation:
hope ths hlp u