*An auto travels at a rate of 25 km/hr for 4 minutes, then at 50 km/hr for 8 minutes, and finally at 20 km/hr for 2 minutes. Find the total distance covered in km and the average speed for the complete trip in m/s.
Answers
v₁ = 25 km/h
t₁ = 4 minutes = (4/60) h
d₁ = distance traveled in 4 min = v₁ t₁ = 25 (4/60) = 1.67 km
v₂ = 50 km/h
t₂ = 8 minutes = (8/60) h
d₂ = distance traveled in 8 min = v₂ t₂ = 50 (8/60) = 6.67 km
v₃ = 20 km/h
t₃ = 2 minutes = (2/60) h
d₃ = distance traveled in 2 min = v₃ t₃ = 20 (2/60) = 0.67 km
d = total distance
Total distance is given as
d = d₁ + d₂ + d₃
d = 1.67 + 6.67 + 0.67
d = 9.01 km
t = total time = t₁ + t₂ + t₃ = 4 + 8 + 2 = 14 min = (14/60) hour
average speed is given as
v = d/t
v = 9.01/(14/60)
v = 38.6 km/h
v = 38.6 (5/18) m/s
v = 10.7 m/s
Answer:
the average speed for the complete trip in m/s is 10.7 m/s
Explanation:
Average pace is calculated through dividing the entire distance that some thing has traveled through the entire quantity of time it took it to journey that distance. Speed is how rapid some thing goes at a specific moment. Average pace measures the common fee of pace over the volume of a trip.The maximum not unusualplace system for common pace is distance traveled divided through time taken. The different system, when you have the preliminary and very last pace, upload the 2 together, and divide through 2
v₁ = 25 km/h
t₁ = four minutes = (four/60) h
d₁ = distance traveled in four min = v₁ t₁ = 25 (four/60) = 1.sixty seven km
v₂ = 50 km/h
t₂ = eight minutes = (eight/60) h
d₂ = distance traveled in eight min = v₂ t₂ = 50 (eight/60) = 6.sixty seven km
v₃ = 20 km/h
t₃ = 2 minutes = (2/60) h
d₃ = distance traveled in 2 min = v₃ t₃ = 20 (2/60) = 0.sixty seven km
d = general distance
Total distance is given as
d = d₁ + d₂ + d₃
d = 1.sixty seven + 6.sixty seven + 0.sixty seven
d = 9.01 km
t = general time = t₁ + t₂ + t₃ = four + eight + 2 = 14 min = (14/60) hour
common pace is given as
v = d/t
v = 9.01/(14/60)
v = 38.6 km/h
v = 38.6 (5/18) m/s
v = 10.7 m/s
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