Running Dogs

Geometry Level 2

Two dogs start running around a 410 m circular track from the same point at the same time in opposite directions and with speeds in the ratio 7:9 . How far from the starting point do they meet for the seventh time?


The answer is 25.625.

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1 solution

Ninad Akolekar
Nov 17, 2014

[HINT]

Just go in the frame of any one dog and find the time of seventh meeting. Now as time remains same in any frame, come back in Earth's frame and check that how much distance any one of the dogs has traveled in that time (You can check this for any dog because both would be at same position, as it is their meeting time. Moreover, as both are at same position, their separation from initial position will also be same, which is the required answer). Subtract 410 from it as many times as you can, as we need only the separation from initial position.

Finally you get answer as 25.625 . I did it this way. Post your solution, if you have a better one.

Based on assumption that speeds of dogs are constant. There're system of equations L1 = V1 * t (running distance till meet for first dog) L2 = V2 * t(running distance till meet for second dog) L = L1 + L2 = 410m a = V1/V2 = 7/9 (v2>v1 and L2>L1) solve system of equations for 1st meeting point: L1 = L/(1 + a) = (410 * 9)/16 = 230.625 m Each time meeting point will be shifted from previous meeting point by 230.625 m. I calculate 7th meeting point absolute distance D7 from D1 = L1. D7 = 7 * D1 = 7 * 230.625 = 1614.375 m. Based of circular track I calculate partial laps: dD7=D7/L=1614.375/410 = 3.9375 times or dD7 = 0.9375 * 410 = 384.375 m or from different point of view 410-384.375=25.625m Answer is either 384.375m or 25.625m

Alex Gawkins - 6 years, 6 months ago

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