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Kinematics

JEE Main level, numerical answer, medium

26 of 308

Question 26 of 308

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 _____ ${ms}^{-1}$ .

Show answer and solution

Answer: 18

Let the total distance be

d

.
Time taken to cover

{\left(\dfrac{1}{3}\right)}^{rd}

of total distance is

{t}_{1}=\dfrac{d}{3{v}_{1}}

.
Time taken to cover

{\left(\dfrac{1}{3}\right)}^{rd}

of total distance is

{t}_{2}=\dfrac{d}{3{v}_{2}}

.
Time taken to cover

{\left(\dfrac{1}{3}\right)}^{rd}

of total distance is

{t}_{3}=\dfrac{d}{3{v}_{3}}

.
Average speed

=\dfrac{Totaldistance}{Totaltime}

{v}_{avg}=\dfrac{d}{\dfrac{d}{3{v}_{1}}+\dfrac{d}{3{v}_{2}}+\dfrac{d}{3{v}_{3}}}

\Rightarrow {v}_{avg}=\dfrac{3\left({v}_{1}{v}_{2}{v}_{3}\right)}{{v}_{1}{v}_{2}+{v}_{2}{v}_{3}+{v}_{3}{v}_{1}}

Here,

{v}_{1}=11m{s}^{-1}

, then,

{v}_{3}=33m{s}^{-1},{v}_{2}=22m{s}^{-1}

{V}_{avg}=\dfrac{3\times 11\times 22\times 33}{\left(11\times 22\right)+\left(22\times 33\right)+\left(33\times 11\right)}=18m{s}^{-1}

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JEE Main 2025 Apr 2 Shift 1 question paper

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