Question

$OA$ and $OB$ are two roads enclosing an angle of ${120}^{°}$ $X$and $Y$ start from $\text{'}O\text{'}$ at the same time. $X$ travels along $OA$ with a speed $4\text{km/hour}$ and $Y$ travels along $OB$ with a speed $3\text{km/hour}$. The rate at which the shortest distance between $X$and $Y$ is increasing after $1\text{hour}$ is

• A. $\sqrt{37}\text{km/h}$
• B. $37\text{km/h}$
• C. $13\text{km/h}$
• D. $\sqrt{13}\text{km/h}$

Answer 1

The correct option is A

$\sqrt{37}\text{km/h}$

Explanation for the correct option:

Find the shortest distance between $X$and $Y$:

Given data,

$X=4\text{km/hour}$

$Y=3\text{km/hour}$

After the time $t$ hours

$X=4t\text{km/hour}$

$Y=3t\text{km/hour}$

Cosine law to find the shortest distance

$c^2=a^2+b^2-2ab\cos c$

Then,

$$\begin{gathered} A^2={\left(4t\right)}^2+{\left(3t\right)}^2-2\left(4t\right)\left(3t\right)\cos \left({120}^{°}\right) \\ =16t^2+9t^2-24t^2\left(\frac{1}{2}\right)\mathbf{}\left[\text{where,}\cos \left({120}^{°}\right)=\frac{1}{2}\right] \\ =16t^2+9t^2-12t^2 \\ =37t^2 \end{gathered}$$

To find the rate at which shortest distance

Differentiate the above equation with respect to $t$

$$\begin{gathered} 2A\frac{dA}{dt}=37\times 2t \\ A\frac{dA}{dt}=37\times t \end{gathered}$$

Put $t=1$

$$\begin{gathered} A\frac{dA}{dt}=37\times 1 \\ A.A\text{'}=37 \end{gathered}$$

When,

$$\begin{gathered} A^2=37\times 1^2 \\ A=\sqrt{37} \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow A.A\text{'}=37 \\ \Rightarrow \sqrt{37},A\text{'}=37 \\ \Rightarrow A\text{'}=\frac{37}{\sqrt{37}} \\ \Rightarrow A\text{'}=\sqrt{37}\text{km/h} \end{gathered}$$

Therefore, the correct answer is option (A).