A car covers AB distance with first one-third at velocity \(\mathrm{v}_{1} \mathrm{~ms}^{-1}\), second one-third at \(\mathrm{v}_{2} \mathrm{~ms}^{-1}\) and last one-third at \(\mathrm{v}_{3} \mathrm{~ms}^{-1}\). If \(\mathrm{v}_{3}=3 \mathrm{v}_{1}, \mathrm{v}_{2}=2 \mathrm{v}_{1}\) and \(\mathrm{v}_{1}=11 \mathrm{~ms}^{-1}\) then the average velocity of the car is ______ \(\mathrm{ms}^{-1}\).

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Kinematics

A car covers AB distance with first one-third at velocity

v_{1} ms^{- 1}

, second one-third at

v_{2} ms^{- 1}

and last one-third at

v_{3} ms^{- 1}

. If

v_{3} = 3 v_{1}, v_{2} = 2 v_{1}

and

v_{1} = 11 ms^{- 1}

then the average velocity of the car is ______

ms^{- 1}

Answer: 18

Average velocity is total displacement divided by total time. Let the total distance be

ℓ

. Time for each segment:

t_{1} = \frac{ℓ /3}{v _{1}}

t_{2} = \frac{ℓ /3}{v _{2}}

t_{3} = \frac{ℓ /3}{v _{3}}

. Total time:

t = \frac{ℓ}{3} (\frac{1}{v _{1}} + \frac{1}{v _{2}} + \frac{1}{v _{3}})

. Average velocity:

v_{avg} = \frac{ℓ}{t} = \frac{3}{\frac{1}{v _{1}} + \frac{1}{v _{2}} + \frac{1}{v _{3}}}

. Substituting

v_{1} = 11

v_{2} = 22

v_{3} = 33

v_{avg} = \frac{3}{\frac{1}{11} + \frac{1}{22} + \frac{1}{33}} = \frac{3}{\frac{6 + 3 + 2}{66}} = \frac{3 \times 66}{11} = 18 m/s

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer