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Two boys are standing at ends $A$ and $B$ of a ground where $A B=200 \mathrm{~m}$. The boy at $B$ starts running in a direction perpendicular to $A B$ with a speed of $6 \mathrm{~ms}^{-1}$. The boy at $A$ starts simultaneously with a velocity of $10 \mathrm{~ms}^{-1}$ and catches the other at time, $t$ where the time, $t$ is
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Verified Answer
The correct answer is:
$25 \mathrm{~s}$
The given situation is shown below
$$
\begin{aligned}
&v_{A}=10 \mathrm{~m} / \mathrm{s} \\
&v_{B}=6 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
The two boys meet at point $C$ after a time $t$. Horizontal component of velocity of $v_{A}$.
$$
\begin{aligned}
v_{A B} &=v_{A} \cos \theta=10 \cos \theta \\
&=10\left(\frac{\sqrt{10^{2}-6^{2}}}{10}\right) \\
&=8 \mathrm{~m} / \mathrm{s} \\
\therefore \quad t &=\frac{A B}{v_{A B}}=\frac{200}{8}=25 \mathrm{~s}
\end{aligned}
$$
$$
\begin{aligned}
&v_{A}=10 \mathrm{~m} / \mathrm{s} \\
&v_{B}=6 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
The two boys meet at point $C$ after a time $t$. Horizontal component of velocity of $v_{A}$.
$$
\begin{aligned}
v_{A B} &=v_{A} \cos \theta=10 \cos \theta \\
&=10\left(\frac{\sqrt{10^{2}-6^{2}}}{10}\right) \\
&=8 \mathrm{~m} / \mathrm{s} \\
\therefore \quad t &=\frac{A B}{v_{A B}}=\frac{200}{8}=25 \mathrm{~s}
\end{aligned}
$$
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