# See question below.Bob can swim at a rate of 5km/h. He is in a river that is flowing at a rate of 9 km/h. a) if bob swims upstream, what is his relative velocity to the ground? b)if bob swims...

See question below.

Bob can swim at a rate of 5km/h. He is in a river that is flowing at a rate of 9 km/h.

a) if bob swims upstream, what is his relative velocity to the ground?

b)if bob swims downstream, what is his relative velocity to the ground?

c) Someone decides to help Bob out of the water with a rope. bob swims across the river (perpendicular to the flow of water) in the same direction he's being pulled. The rescuer is pulling at a rate of 3 km/h. What is Bob's relative velocity to the ground?

### 1 Answer | Add Yours

Well, as far as swimming upstream, they call it upstream for a reason; it's definitely harder to swim upstream, depending on the river flow! If the river is flowing at a velocity of 9 km/hr, and Bob is swimming at a rate of 5 km/hr, he is losing the battle with the river and is going downstream at a rate of 4 km/hr. You would simply subtract the two rates to get the net result, in the direction of the river, since it was the larger number. If Bob swims downstream, you would do the opposite, add the two rates together, so the rivers 9 km/hr plus Bobs 5 km/hr would equal 14 km/hr, with respect to the ground.

Part (c) is a little harder. If Bobs friend is pulling at a rate of 3 km/hr, that added to Bobs rate of 5 km/hr equals 8 km/hr. The river is flowing at 9 km/hr. If you draw this out like the legs of a triangle, with one leg being 9 inches, the other being 8 inches, and then measure the hypotenuse of the triangle. it measures 12 inches. That would indicate a velocity of 12 km/hr, with respect to the ground. Another way to figure this out would be to use the formula a2 + b2 = c2, where the square of one leg plus the square of the other leg equals the square of the hypotenuse. 9 squared is 81, plus 8 squared, which is 64, equals 145. The square root of 145 is 12.04.

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