alberto drove to the lake and back. the trip there took two hours and the trip back took three hours. he averaged 20 km/h faster on the trip there than on the return trip. What was alberto's average speed on the outbound trip?
speed on first trip --- x km/h
speed on return trip --- x-20
let the distance be d
time for first part = d/x
time for 2nd part = d/(x-20)
d/x = 2 ---> d = 2x
d/(x-20) = 3 ---> d = 3x - 60
3x-60 = 2x
x = 60
So the distance each way was 120 km
average speed = 240/5 = 48 km/h
Notice that the two speeds were 60 km/h and 40 km/h, and to get the average speed we CANNOT just average the two speeds.
since distance = speed * time, if the outbound speed was x, then
2x = 3(x-20)
To find Alberto's average speed on the outbound trip, we need to determine his distance traveled and the time it took him.
Let's assume his average speed on the return trip (trip back) is "x" km/h. Since the trip there took him two hours and the trip back took him three hours, we know the distances traveled are the same, as it is a round trip.
Using the formula:
Distance = Speed × Time
For the trip there, the distance traveled is equal to the speed multiplied by the time:
Distance = (x + 20) km/h × 2 hours
And for the trip back, the distance traveled is equal to the speed multiplied by the time:
Distance = x km/h × 3 hours
Since both distances are the same, we can set them equal to each other:
(x + 20) km/h × 2 hours = x km/h × 3 hours
Simplifying the equation:
2(x + 20) = 3x
Expanding the brackets:
2x + 40 = 3x
Rearranging the equation:
3x - 2x = 40
x = 40
Therefore, Alberto's average speed on the outbound trip (trip there) is 40 km/h.