A person travelling on a straight line moves with a uniform velocity v_1 for a distance x and with a uniform velocity v_2 for the next (3)/(2) x distance. The average velocity in this motion is (50)/(7) ~m / s . If v_1

Correct answer: 0 aligned & v_ avg = x_1+x_2 t_1+t_2 & (50)/(7)= x+ 3 x 2 x 5 + 3 x 2 v_2 & (50)/(7)= 5 / 2 (1)/(5)+ 3 2 v_2 & (1)/(5)+ 3 2 v_2 =(7)/(20) aligned aligned & 3 2 v_2 =(7)/(20)-(1)/(5)=(7-4)/(20) & 3 2 v_2 =(3)/(20) & v_2=10 m / s aligned

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Physics
Chapter: Motion in One Dimension

JEE Main 2025 · 2 April · Morning Shift · Physics · Motion in One Dimension

A person travelling on a straight line moves with a uniform...

A person travelling on a straight line moves with a uniform velocity

\mathrm{v}_1

for a distance x and with a uniform velocity

\mathrm{v}_2

for the next

\frac{3}{2} \mathrm{x}

distance. The average velocity in this motion is

\frac{50}{7} \mathrm{~m} / \mathrm{s}

. If

\mathrm{v}_1

5 \mathrm{~m} / \mathrm{s}

then

\mathrm{v}_2=

_________

\mathrm{m} / \mathrm{s}

Correct answer

Step-by-step solution

\begin{aligned} & \mathrm{v}_{\text{avg }}=\frac{\mathrm{x}_1+\mathrm{x}_2}{\mathrm{t}_1+\mathrm{t}_2} \\ & \Rightarrow \frac{50}{7}=\frac{\mathrm{x}+\frac{3 \mathrm{x}}{2}}{\frac{\mathrm{x}}{5}+\frac{3 \mathrm{x}}{2 \mathrm{v}_2}} \\ & \Rightarrow \frac{50}{7}=\frac{5 / 2}{\frac{1}{5}+\frac{3}{2 \mathrm{v}_2}} \\ & \Rightarrow \frac{1}{5}+\frac{3}{2 \mathrm{v}_2}=\frac{7}{20}\end{aligned}

\begin{aligned} & \Rightarrow \frac{3}{2 \mathrm{v}_2}=\frac{7}{20}-\frac{1}{5}=\frac{7-4}{20} \\ & \Rightarrow \frac{3}{2 \mathrm{v}_2}=\frac{3}{20} \\ & \Rightarrow \mathrm{v}_2=10 \mathrm{ m} / \mathrm{s}\end{aligned}

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