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Question:

Sprinter A traverses distances A₁A₂, A₂A₃, and A₃A₁ at average speeds of 20, 30 and 15 respectively. B traverses her entire path at a uniform speed of $(10\sqrt{3}+20).$ C traverses distances C₁C₂, C₂C₃, and C₃C₁ at average speeds of $\frac{40}{3}(\sqrt{3}+1),\frac{40}{3}(\sqrt{3}+1),$ and 120 respectively. All speeds are in the same unit. Where would B and C be respectively when A finishes her sprint?

Explanation:

Time taken by A to travel through distance a (A₁A₂ + A₂A₃ + A₃A₁)

$=\frac{2r}{20}+\frac{2r}{30}+\frac{2r}{15}=\frac{3r}{10}$

Distance travelled by B in the same time,

$=\frac{3r}{10}\times (10\sqrt{3}+20)=3r(2+\sqrt{3})=b$

∴ B travels a full round in the same time as A. Thus, B will be at B₁.

For C, time taken to traverse through C₁C₂

_{$=\frac{2r(\sqrt{3}+1)}{{\displaystyle \frac{40}{3}}(\sqrt{3}+1)}=\frac{3r}{20}$}

Time remaining for C = $=\left(\frac{3}{10}-\frac{3}{20}\right)r=\frac{3r}{20}$

Now, distance travelled by C in this time = $\frac{3r}{20}\times \frac{40}{32}(1+\sqrt{3})=2r(1+\sqrt{3})=C_2{C}_3$

∴ C will be at C₃.

Hence, option (c).