In a canoe race, a team paddles downstream 480 m in 60 seconds. the same team makes the trip upstream and 80 seconds. find the teammates rate in stillwater and the rate of the current period

The rate in still water is 8m/s

The rate in current period is is 6 m/s

Step-by-step explanation:

In a canoe race, A team paddles downstream 480 m in 60 seconds. The same team makes the trip upstream and 80 seconds.

We observe that it took the team more time paddling upstream than paddling downstream even though it was the same distance.

Let us assume that on paddling downstream, they paddled in the same direction with the current. This means that they paddled on still water. On paddling upstream, they paddled in the opposite direction of the current.

Let the speed of the boat or teammates be

x m/s

Let the speed of the current be

y m/s

Distance = speed × time

Distance travelled on still water or downstream

= (x+y) × 60 = 60(x+y)

Distance travelled on upstream

= (x-y) × 80 = 80(x-y)

Since the distance is 480 miles for both upstream and downstream,

60(x+y) = 480

x + y = 480/60 = 8 - - - - - -1

80(x-y) = 480

x - y = 480/80 = 6 - - - - - -2

Adding equation 1 and 2,it becomes

2x = 14

x = 14/2 = 7 m/s

y = 8 - x = 8-7

y = 1 m/s

Rate in still water = x +y = 7+1 = 8m/s

Rate in current period = x - y = 7 - 1 = 6m/s

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