JEE Main 2023 · 25 January · morning · Single correct · +4 / −1
A car travels a distance of 'x' with speed v₁ and then same distance 'x' with speed v₂ in the same direction. The average speed of the car is :
- (a) v₁v₂/2(v₁ + v₂)
- (b) 2v₁v₂/(v₁ + v₂)
- (c) 2x/(v₁ + v₂)
- (d) (v₁ + v₂)/2
Answer and solution
Answer (b) 2v₁v₂/(v₁ + v₂)
Solution
Physics sir · 3 steps
Average speed is total distance over total time, so find the time for each half separately first.
- Time for each part. The first stretch takes t₁ = x/v₁ and the second takes t₂ = x/v₂. The total distance covered is 2x.
v_(avg) = 2x/(x/v₁ + x/v₂) = 2x/x((v₁ + v₂)/(v₁ v₂))
- Cancel x. The x cancels from top and bottom, leaving v_(avg) = (2 v₁ v₂)/(v₁ + v₂). This is the harmonic mean of the two speeds.
- So. The average speed is (2 v₁ v₂)/(v₁ + v₂), which is option (b).
The idea to keep. Equal distances at two speeds give the harmonic mean, 2v₁v₂/(v₁+v₂). Equal times give the plain average. Always ask which one is equal before you average.
Answer from the NTA final answer key. Solution by PrepWiser.