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 time interval 0 to 50 min.
Answers
Answered by
53
Hii dear,
# Answers-
1. Magnitude of average velocity = 0 km/h
2. Average speed = 6 km/h
# Explaination-
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
Total distance travelled = 2.5+2.5 = 5 km.
Displacement = 0 km
# Formula-
1. Magnitude of average velocity
= displacement/time
= 0/50 min
= 0 km/h
2. Average speed
= total distance travelled/time
= 5/50
= 1/10 km/min
= 1/10×60 = 6 km/h
Hope you got your answer...
# Answers-
1. Magnitude of average velocity = 0 km/h
2. Average speed = 6 km/h
# Explaination-
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
Total distance travelled = 2.5+2.5 = 5 km.
Displacement = 0 km
# Formula-
1. Magnitude of average velocity
= displacement/time
= 0/50 min
= 0 km/h
2. Average speed
= total distance travelled/time
= 5/50
= 1/10 km/min
= 1/10×60 = 6 km/h
Hope you got your answer...
Answered by
19
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