Question

Ship moving with velocity${\overrightarrow{V}}_1=30\stackrel{⏜}{i}+50\stackrel{⏜}{j}$ from position $(0,0)$ and ship B moving with velocity ${\overrightarrow{V}}_2=-10\stackrel{⏜}{i}$ from position $(80,150)$. The time for minimum separation is

• A. $2.6$
• B. $2.2$
• C. $2.4$
• D. None of these

Answer 1

The correct option is A

$2.6$

Step 1: Given data:

Velocity of ship A$\left({\overrightarrow{V}}_1\right)=30\stackrel{⏜}{i}+50\stackrel{⏜}{j}$

Velocity of ship B$\left({\overrightarrow{V}}_2\right)=-10\stackrel{⏜}{i}$

Step 2: Formula used:

The time for minimum separation is given by-

$t_{min}=\frac{\left|V_r.r_r\right|}{{\left|V_r\right|}^2}\dots \dots \dots \dots \dots \left(1\right)$

Where, $V_r$ is resultant velocity, $r_r$is the position vector.

Step 3: Calculation of time for minimum separation

It is given that

$$\begin{gathered} \begin{array}{l}{V}_1=30\hat{i}+50\hat{j}\end{array} \\ \begin{array}{l}{V}_2=-10\hat{i}\end{array} \end{gathered}$$

The resultant $\begin{array}{l}{V}_r=V_1-V_2=40i+50j\end{array}$

Relative distance $\begin{array}{l}{r}_r=-80\hat{i}-150\hat{j}\end{array}$

$\begin{array}{rcl}{V}_r{r}_r& =& \left(40i+50j\right)\left(-80i-150j\right)\\ & =& -3200-7500\\ & =& -10700\dots \dots \dots \left(2\right)\end{array}$

$\begin{array}{rcl}{V}_r& =& \sqrt{{40}^2+{50}^2}\\ & =& 64.03\dots \dots \dots \dots \dots \left(3\right)\end{array}$

The time for minimum separation is using the formula in equation (1)

Substitute the values from equation (2) and (3)

$\begin{array}{rcl}\begin{array}{l}\frac{\left|V_r.r_r\right|}{{\left|V_r\right|}^2}=\frac{\left|-10700\right|}{{\left|64.03\right|}^2}\end{array}& =& \frac{10700}{4100}=2.6s\\ & & \end{array}$

Hence option A is the correct answer.