AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

The figure below represents the ships A and B. The time required to travel by ship A from A to C is, $t=\;\frac{x_{A}}{v_{A}}$ $t=\;\frac{x_{A}}{20}$ The time required to travel by ship $B$ is, $t=\;\frac{x_{B}}{v_{B}}$ $t=\;\frac{x_{B}}{10}$ From above two equations, $\;\frac{x_{A}}{20}=\;\frac{x_{B}}{10}$ $x_{A}=2x_{B}$ The shortest distance is, $AB^{2}=x_{A}^{2}+(200-x_{2}^{B})^{2}$ $AB=\sqrt{(2x_{A})^{2}+(200-x_{2}^{B})^{2}}$ $AB=\sqrt{5x_{B}^{2}-400x_{B}+40000}\;\;\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\;(\mathrm{I})$ For the shortest distance, $\;\frac{d}{dx_{B}}(AB)=0$ $0=\;\frac{1}{2\sqrt{5x_{B}^{2}-400x_{B}+40000}}(10x_{B}-400)\;\;\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\;(\mathrm{II})$ $x_{B}=40\,\text{m}$Again differentiate equation (II). $\;\frac{d^{2}}{dx_{B}^{2}}(AB)>0_{\text{at}\;x=40\;\text{m}}$ The distance always greater than zero Substitute the value in equation (I). $AB=\sqrt{5\times40^{2}-400\times40+40000}$ $AB=80\sqrt{5}\,\text{km}$ The time after which the distance between the ships is shortest is, $t=\;\frac{x_{B}}{10}=\;\frac{40}{10}=4\,\text{hr}$