A car covers AB distance with first one-third at velocity v₁ ms⁻¹, second one-third at v₂ ms⁻¹ and last one-third at v₃ ms⁻¹. If v₃ = 3v₁, v₂ = 2v₁ and v₁ = 11 ms⁻¹ then the average velocity of the car is _____________ ms⁻¹.

Solution

Physics sir · 3 steps

Equal distances, different speeds: add up the time for each third, then divide total distance by total time.

1. Set up times. Let each third of AB be d, so the total distance is 3d. The car spends d/v₁, d/v₂ and d/v₃ on the three parts.

   v_(avg) = 3d/(d/v₁+d/v₂+d/v₃) = 3/(1/v₁+1/v₂+1/v₃)

2. Put in numbers. Here v₁ = 11, v₂ = 22 and v₃ = 33 m s⁻¹. Then 1/11+1/22+1/33 = (6+3+2)/66 = 11/66 = 1/6.
3. So. v_(avg) = 3/(1/6) = 18 m s⁻¹. The answer is 18.

The common slip: 22. Taking the plain average (11+22+33)/3 gives 22, not 18. That works only for equal time intervals, not equal distances.

The idea to keep. Average velocity is always total displacement over total time. For equal distances, it is the harmonic mean of the speeds, not the ordinary average.

Answer from the NTA final answer key. Solution by PrepWiser.