A man walks on a straight road from his home to a market 2.5 km away
with a speed of 15km /h. Finding the market closed, he instantly turns and
walks back home with a speed of 17.5km/ h. What is the (a) Magnitude of
Average Velocity and (b) Average Speed of the man over the time interval 0
to 50 min?
Answers
Answer:
Speed = 6 km/h
Speed = 6 km/hDistance = x km
Speed = 6 km/hDistance = x kmTime = Distance/Speed
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/h
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x km
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x km
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x kmTotal Time taken = x/6 + x/2
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x kmTotal Time taken = x/6 + x/2= (x + 3x)/6
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x kmTotal Time taken = x/6 + x/2= (x + 3x)/6= 4x/6
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x kmTotal Time taken = x/6 + x/2= (x + 3x)/6= 4x/6= (2x/3) hrs.
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x kmTotal Time taken = x/6 + x/2= (x + 3x)/6= 4x/6= (2x/3) hrs.Avreage speed of boy = Total Distance covered / Total Time taken
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x kmTotal Time taken = x/6 + x/2= (x + 3x)/6= 4x/6= (2x/3) hrs.Avreage speed of boy = Total Distance covered / Total Time taken= 2x/2x/3
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x kmTotal Time taken = x/6 + x/2= (x + 3x)/6= 4x/6= (2x/3) hrs.Avreage speed of boy = Total Distance covered / Total Time taken= 2x/2x/3= 2x * 3/2x
Speed = 6 km/hDistance = x kmTime = Distance/Speed= (x/6) hrs.Case 2Speed = 2 km/hDistance = x kmTime = Distance/Speed= (x/2) hrs.Now, Total Distance covered by boy = 2x kmTotal Time taken = x/6 + x/2= (x + 3x)/6= 4x/6= (2x/3) hrs.Avreage speed of boy = Total Distance covered / Total Time taken= 2x/2x/3= 2x * 3/2x= 3 km/h
Answer:
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