Question

Pat, Shannon and Fido (their dog) are on a circular path 100 feet around. Fido is very excited. Shannon starts walking around the path maintaining a speed of 1 foot per second, while Pat stays still. Fido begins racing back and forth between Pat and Shannon on the circular path along the section of the path Shannon has NOT walked on. What is Fido’s average speed if he has just returned to Pat for the 5th time when Shannon gets to the half-way point? **IMPORTANT** Assume Fido maintains a constant speed of 3 feet per second for this last run back to Pat. **IMPORTANT** Also, assume that Fido met Shannon at equally spaced moments in time (including the starting).

Answer 1

perimeter = 100 feet

(s stands for speed)

F meets S at equal intervals of time, call it seconds. Since S is walking at 1 fps, the distance covered by S in seconds is feet.

S has walked halfway (50 feet) when F meets for 5th time.

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So, P is stationary. S starts walking Counter Clockwise (say) and F starts running Clockwise to meet S.

F meets S in seconds

By the time F meets S for the first time, S would have walked a distance feet

Thus, the distance covered by F is 100 - .

Now, F starts running back. It covers 100 - when it reaches P.

Again it starts and meets S , who has covered 2 feet totally. So, distance F has run this time is 100 - 2. This is the 2nd meeting

The distance for this meeting is thus (100 - +100 - 2)

Repeating the pattern, when F meets third time, the distance for this meeting is

(100 - 2+100 - 3)

The distance for 4th meeting is (100 - 3+100 - 4)

The distance for the 5th meeting is (100 - 4+100 - 5)

But the 5th meeting is covered at halfway point , which is 50 feet. So, now we have a chance to write an equation.

Now, the speed of F during the last run is given as 3 fps. Let us find out if it matches our calculations:

which is correct.

Now, let us determine the speeds for each meeting.

1:

2:

3:

4:

5: 3 fps

The average is thus: