George and Bill left town A for town B at the same time. George was riding a bicycle at a constant speed, while Bill was driving a car. Bill traveled 5 times faster than George. Bill's car broke halfway, and the rest of the way to the town B Bill traveled by foot with a speed equal to half of the George's bicycle speed. Who got to B first: George or Bill?
George and Bill left town A for town B at the same time.
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George Speed - X
Bills Speed - 5X
Bills speed by foot - X/2
d- distance from A-B
X = d/t1 5X + X/2 =d/t2
11X/2 = d/ t2
t1= d/X ---------------- time it takes Bill to go from A to B
t2 = d/ (11X/2)---------- Time it will take George to go from A to B
t1 > t2 , Therefore Bill reaches earlier.
Note : The formula used was
Speed * ( time it would take) = distance traveled.
Let D be the distance between A and B. Let also George's speed be V. Then Bill's car speed was 5V, his walking speed was V/2.
Time it took George to get to B:
time it took Bill to get to B:
It is immediately obvious that, since George's time is L/V, it took Bill longer to get to B. In fact, it took him 11/10*(L/V)=11/10*tG, where tG is the time it took George to get from A to B.
So George got to B first.
The time George completed the trip: t = AB/v
The time Bill completed the trip: t(total) = t1 + t2
The time for the first half of the trip (car is not broken): t1 = (AB/2) / 5v = AB/10v
The time for the second half (by foot): t2 = (AB/2) / (v/2) = AB/v
Total time for Bill : T = t1 + t2 = AB/10v + AB/v. This time is longer than George's time ( AB/v)
Answer: George got to B first