Question 809938: A woman on foot leaves a restaurant heading west at noon. Three hours later, a man on a bicycle leaves the resturant pedaling west at a rate of 12 miles per hour. at six o clock that evening, a woman on a scooter leaves the resturant driving west at a speed of 39 miles per hour. if the first woman was walking at a rate of 3 miles per hour, calculate the distances below.
A. how far from the resturant will the woman on foot be when the man on the bicycle reaches her?
B. How long will it take for the woman on the scooter to reach the woman on foot?
C. How far from the resturant will the man on the bicycle be when the woman on the scooter reaches him
Answer by devipriya(2)   (Show Source):
You can put this solution on YOUR website! Part 1:
Let the time be measured as 't'.
The distance traveled when the woman meets the cyclist will be the same (d=d). The time for the cyclist, however, will be three hours lass than the time for the woman.
Distance = Rate * Time
For the woman, d = 3 * t.
For the cyclist, d = 12 * (t - 3).
The distance is equal, so:
3t = 12(t-3),
3t = 12t - 36,
36 = 9t,
t = 4.
That's twelve miles.
Part 2:
Similar to part 1:
d = 3t,
d = 39(t - 6);
so
3t = 39(t - 6),
3t = 39t - 234,
234 = 36t,
t = 6.5 (hours), that's 19.5 miles away when they meet!
Part 3:
The woman on foot no longer needs considered- the offset of the man-on-bicycle and woman-on-scooter is three hours:
d = 12t,
d = 39(t - 3); so,
12t = 39(t - 3),
12t = 39t - 117,
117 = 27t,
so the time for the man-on-bicycle is 4.333 hours (the woman-on-scooter takes 1.333 hours).
Either way, the distance is 52 miles.
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