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Particle A (which was located at the origin at time $t=0$ ) is moving along the $\mathrm{x}$-axis with a constant speed of $1 \mathrm{~m} / \mathrm{s}$. Location of particle $\mathrm{B}$ which is moving along the $\mathrm{y}$-axis is given by $\mathrm{y}=\mathrm{ct}^2$, where $\mathrm{c}=1 \mathrm{~m} / \mathrm{s}^2$. Find the speed of particle A relative to particle $\mathrm{B}$ at $\mathrm{t}=1 \mathrm{sec}$.
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Verified Answer
The correct answer is:
$\sqrt{5} \mathrm{~m} / \mathrm{s}$
$\begin{aligned}
& \text { For particle along y-axis } \\
& \mathrm{y}=\mathrm{ct}^2 \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{ct} \\
& \Rightarrow \mathrm{V}_{\mathrm{B}}=2 \mathrm{ct} \\
& \text { at } \mathrm{t}=1 \mathrm{sec}, \\
& \overrightarrow{\mathrm{V}}_{\mathrm{B}} 2 \times 1 \times=1 \hat{\mathrm{j}}=2 \hat{\mathrm{j}}
\end{aligned}$
So, $\overrightarrow{\mathrm{V}}_{\mathrm{AB}}=\overrightarrow{\mathrm{V}}_{\mathrm{A}}-\overrightarrow{\mathrm{V}}_{\mathrm{B}}$
$\begin{aligned}
& =\hat{\mathrm{i}}-2 \hat{\mathrm{j}} \\
& \left|\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\right|=\sqrt{1+(-2)^2}=\sqrt{5} \mathrm{~m} / \mathrm{s}
\end{aligned}$
& \text { For particle along y-axis } \\
& \mathrm{y}=\mathrm{ct}^2 \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{ct} \\
& \Rightarrow \mathrm{V}_{\mathrm{B}}=2 \mathrm{ct} \\
& \text { at } \mathrm{t}=1 \mathrm{sec}, \\
& \overrightarrow{\mathrm{V}}_{\mathrm{B}} 2 \times 1 \times=1 \hat{\mathrm{j}}=2 \hat{\mathrm{j}}
\end{aligned}$
So, $\overrightarrow{\mathrm{V}}_{\mathrm{AB}}=\overrightarrow{\mathrm{V}}_{\mathrm{A}}-\overrightarrow{\mathrm{V}}_{\mathrm{B}}$
$\begin{aligned}
& =\hat{\mathrm{i}}-2 \hat{\mathrm{j}} \\
& \left|\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\right|=\sqrt{1+(-2)^2}=\sqrt{5} \mathrm{~m} / \mathrm{s}
\end{aligned}$
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