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Question

Physics Question on Motion in a straight line

A bus travelling the first one-third distance at a speed of 10kmh1{10 \, km \, h^{-1}}, the next one-third at 20kmh1{20 \, km \, h^{-1}} and last one third at 60kmh1{60 \, km \, h^{-1}}. The average speed of bus is

A

16kmh1{16 \, km \, h^{-1}}

B

18kmh1{18 \, km \, h^{-1}}

C

9kmh1{9 \, km \, h^{-1}}

D

48kmh1{48 \, km \, h^{-1}}

Answer

18kmh1{18 \, km \, h^{-1}}

Explanation

Solution

Here v1=10kmh1for1st1/3rddistancev_1 = {10\, km\, h^{-1} for \, 1^{st} \, 1/3^{rd} \, distance}
v2=20kmh1for2nd1/3rddistancev_2 = {20\, km\, h^{-1} for \, 2^{nd} \, 1/3^{rd} \, distance}
v3=60kmh1for3rd1/3rddistancev_3 = {60\, km\, h^{-1} for \, 3^{rd} \, 1/3^{rd} \, distance}
vavg=3v1v2v3v1v2+v2v3+v3v1v_{avg} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}
=3×10×20×6010×20+20×60+60×10= \frac{3 \times 10 \times 20 \times 60}{10 \times 20 + 20 \times 60 + 60\times 10}
vavg=18kmh1v_{avg} ={18 \, km \, h^{-1}}