Walking at 4km/h, Bruce can make the roundtrip between his campsite and Lookout Point in 2.5 h. Rowing on Crooked River, he can row upstream from the campsite to Lookout Point in 1 h and can row back again in 40 min. Find Bruce’s rate of rowing in still water and the speed of the current in Crooked River.
t(avg) = (40 + 60) / 2 = 50min = Time required to row the distance in still water.
V = d / t = 5km / 50min = 0.1km/min = = 6km/h in still water.
V = Vmax - Vmin,
V=5km/(2/3)h -5km / 1h=7.5 -5=2.5km/h = speed of stream.
To find Bruce's rate of rowing in still water and the speed of the current in Crooked River, let's carefully analyze the information provided.
Let's assume Bruce's rate of rowing in still water is represented by 'r' km/h, and the speed of the current is represented by 'c' km/h.
First, let's consider the round trip between the campsite and Lookout Point by walking. Bruce walks at a speed of 4 km/h. The total time taken for this round trip is 2.5 hours.
Since the distance going to and returning from Lookout Point is the same, we can equate the time taken for both journeys:
Time taken to go to Lookout Point + Time taken to return from Lookout Point = Total time taken for the round trip
Distance / Walking speed + Distance / Walking speed = 2.5 hours
Since the distance is the same, we can simplify the equation:
2 * Distance / 4 km/h = 2.5 hours
Distance / 4 km/h = 1.25 hours
Solving for the distance:
Distance = 1.25 hours * 4 km/h
Distance = 5 km
Next, let's consider the round trip between the campsite and Lookout Point by rowing. Bruce takes 1 hour to row upstream from the campsite to Lookout Point and 40 minutes (which is equivalent to 2/3 of an hour) to row downstream from Lookout Point back to the campsite.
When rowing upstream, Bruce's effective speed is reduced by the speed of the current. So, his effective speed is (r - c) km/h.
When rowing downstream, Bruce's effective speed is increased by the speed of the current. So, his effective speed is (r + c) km/h.
The distance between the campsite and Lookout Point is still 5 km.
Using the formula: Time = Distance / Speed, we can set up two equations:
Upstream: 1 hour = 5 km / (r - c) km/h
Downstream: 2/3 hour = 5 km / (r + c) km/h
Now, we have a system of equations to solve:
1) 1 = 5 / (r - c)
2) 2/3 = 5 / (r + c)
To eliminate fractions, we can cross-multiply:
1) (r - c) = 5 / 1
2) (r + c) = 5 / (2/3)
Simplifying the second equation:
2) (r + c) = 5 * (3/2)
2) (r + c) = 15/2
Now we have a system of two equations:
1) r - c = 5
2) r + c = 15/2
Adding the two equations:
1) r - c + r + c = 5 + 15/2
2r = 10 + 15/2
2r = 20/2 + 15/2
2r = 35/2
r = (35/2) / 2
r = 35/4
r = 8.75 km/h
Therefore, Bruce's rate of rowing in still water is 8.75 km/h.
Substituting the value of 'r' into equation 1 to solve for 'c':
r - c = 5
8.75 - c = 5
-c = 5 - 8.75
-c = -3.75
c = 3.75 km/h
Therefore, the speed of the current in Crooked River is 3.75 km/h.
So, Bruce's rate of rowing in still water is 8.75 km/h and the speed of the current in Crooked River is 3.75 km/h.