Kinematics

JEE Main level, numerical answer, medium

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Question 5 of 308

A person travelling on a straight line moves with a uniform velocity v1\mathrm{v}_1 for a distance x and with a uniform velocity v2\mathrm{v}_2 for the next 32x\frac{3}{2} \mathrm{x} distance. The average velocity in this motion is 507 m/s\frac{50}{7} \mathrm{~m} / \mathrm{s}. If v1\mathrm{v}_1 is 5 m/s5 \mathrm{~m} / \mathrm{s} then v2=\mathrm{v}_2= _________ m/s\mathrm{m} / \mathrm{s}.

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Answer: 10

vavg =x1+x2t1+t2⇒507=x+3x2x5+3x2v2⇒507=5/215+32v2⇒15+32v2=720\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}
⇒32v2=720−15=7−420⇒32v2=320⇒v2=10 m/s\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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