Pilar is playing with a remote controlled toy boat in a lake. She navigates the boat 400 m away from herself, keeping it at a constant speed. Steering it back toward herself, Pilar navigates the boat over the same route and at the same speed for 2 minutes, and then she increases the boat’s speed by 10m/min. The return trip is 60 seconds faster. How long does the return trip take?
Let's break down the information provided:
- Distance from Pilar to the boat: 400m
- Speed of the boat during the initial trip: let's call it X m/min
- Time taken for the initial trip: unknown, let's call it T min
- Speed of the boat after increasing speed: X + 10 m/min
- Time taken for the return trip: T - 60 seconds
During the initial trip, the boat traveled 400m at speed X m/min for time T min, so we have the equation:
400 = X * T
During the return trip, the boat traveled the same 400m at speed X + 10 m/min for time T - 60 seconds (which is T - 1 min), so we have the equation:
400 = (X + 10) * (T - 1)
Now, we can solve the two equations simultaneously:
From the first equation, we get T = 400 / X
Substitute T into the second equation:
400 = (X + 10) * (400 / X - 1)
Expanding the equation:
400 = 400 + 4000 / X - X - 10
Combine like terms:
4000 = 4000 / X - X - 10
Add X to both sides:
4000 + X = 4000 / X - 10
Multiply both sides by X:
4000X + X^2 = 4000 - 10X
Rearrange the equation:
X^2 + 4010X - 4000 = 0
Now, we solve for X using the quadratic formula:
X = (-4010 ± sqrt(4010^2 - 4*1*(-4000))) / 2*1
X = (-4010 ± sqrt(16080100 + 16000)) / 2
X = (-4010 ± sqrt(16096100)) / 2
X = (-4010 ± 4010) / 2
X = (-8010 / 2) or X = (0 / 2)
X = -4005 or X = 0 - since speed cannot be negative, the speed of the boat during the initial trip is 0.
Therefore, the boat was not moving during the initial trip, and Pilar could not travel 400m away from herself by the boat. The return trip would not have been possible.