An aeroplane flies around a square field ABCD of each side 1000 km . Its speed along AB is $250 \mat

Question

An aeroplane flies around a square field ABCD of each side 1000 km . Its speed along AB is $250 \mathrm{kmh}^{-1}$, along BC $500 \mathrm{kmh}^{-1}$, along CD $200 \mathrm{kmh}^{1}$, and along DA $100 \mathrm{kmh}^{-1}$. Its average speed (in $\mathrm{kmh}^{-1}$ ) over the entire trip is\

Professor PC Thomas Classes · Answer

\ Let $t_{A B} \cdot t_{B C}, t_{C D}$ and $t_{D A}$ be the time taken by the aeroplane to go from A to $\mathrm{B}, \mathrm{B}$ to $\mathrm{C}, \mathrm{C}$ to D and D to A respectively.\ $	herefore \mathrm{t}_{\mathrm{AB}}=\frac{1000 \mathrm{~km}}{250 \mathrm{kmh}^{-1}}=4 \mathrm{~h}$\ $	herefore \mathrm{t}_{\mathrm{BC}}=\frac{100 \mathrm{~km}}{500 \mathrm{kmh}^{-1}}=2 \mathrm{~h}$\ $	herefore \mathrm{t}_{\mathrm{CD}}=\frac{1000 \mathrm{~km}}{200 \mathrm{kmh}^{-1}}=5 \mathrm{~h}$\ $	herefore \mathrm{t}_{\mathrm{DA}}=\frac{100 \mathrm{~km}}{100 \mathrm{kmh}^{-1}}=10 \mathrm{~h}$\ Average speed $=\frac{	ext { Total distance covered }}{	ext { Total time taken }}$\ $=\frac{1000 \mathrm{~km}+1000 \mathrm{~km}+1000 \mathrm{~km}+1000 \mathrm{~km}}{\mathrm{t}_{\mathrm{AB}}+\mathrm{t}_{\mathrm{BC}}+\mathrm{t}_{\mathrm{CD}}+\mathrm{t}_{\mathrm{DA}}}$\ $=\frac{4000 \mathrm{~km}}{4 \mathrm{~h}+2 \mathrm{~h}+5 \mathrm{~h}+10 \mathrm{~h}}=\frac{4000 \mathrm{~km}}{21 \mathrm{~h}}=190.5 \mathrm{kmh}^{-1}$\