Physics, asked by shdy42, 1 year ago

Solve the following
1) A moving body travels distance d at uniform speed v1, and travel distance d at uniform speed v2,prove that its
average speed is 2v1v2/2
V + v2​

Answers

Answered by Anonymous
1

hope it helps..........

Attachments:
Answered by amitnrw
1

Answer:

2v₁ v₂ / (v₁ + v₂)

Explanation:

A moving body travels distance d at uniform speed v1, and travel distance d at uniform speed v2,

Distance = d

Uniform speed = v₁

Time = Distance / Speed

=> Time = d/v₁

Distance = d

Uniform speed = v₂

Time = Distance / Speed

=> Time = d/v₂

Total distance = d + d = 2d

Total time = d/v₁ + d/v₂ = d(1/v₁ + 1/v₂)

= d(v₂ + v₁)/(v₁ v₂)

Average speed = Total distance / Total time

= 2d  / (d(v₂ + v₁)/(v₁ v₂))

=  2v₁ v₂ / (v₂ + v₁)

=  2v₁ v₂ / (v₁ + v₂)

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