JEE Main 2023 · 1 February · morning · Single correct · +4 / −1
An object moves with speed v₁,v₂ and v₃ along a line segment AB, BC and CD respectively as shown in figure. Where AB = BC and AD = 3AB, then average speed of the object will be:
- (a) v₁v₂v₃/3(v₁v₂ + v₂v₃ + v₃v₁)
- (b) (v₁ + v₂ + v₃)/3
- (c) (v₁ + v₂ + v₃)/3v₁v₂v₃
- (d) 3v₁v₂v₃/(v₁v₂ + v₂v₃ + v₃v₁)
Answer and solution
Answer (d) 3v₁v₂v₃/(v₁v₂ + v₂v₃ + v₃v₁)
Solution
Physics sir · 3 steps
Equal distances, not equal times: add up the time on each piece, then divide total distance by it.
- Find each segment. Let AB = x. Then BC = x, and since AD = 3x, the last piece is CD = 3x - x - x = x. All three segments are equal.
- Time on each segment. Time is distance over speed, so the total time is x/v₁ + x/v₂ + x/v₃. Taking the common denominator gives x (v₂ v₃ + v₁ v₃ + v₁ v₂)/(v₁ v₂ v₃).
v_(avg) = 3x/x(1/v₁+1/v₂+1/v₃) = (3v₁ v₂ v₃)/(v₁ v₂ + v₂ v₃ + v₃ v₁)
- So. The x cancels, and the average speed is the harmonic mean of the three speeds. That matches (d).
The idea to keep. Average speed is always total distance over total time. Equal distances give the harmonic mean of the speeds. Only equal times give the plain arithmetic mean.
Answer from the NTA final answer key. Solution by PrepWiser.