Question

Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you fall out of your boat and immediately grab a piling of the Warm Springs Bridge. You hold on for 40 s and then swim after the boat with a speed relative to the water of 0.95 m/s. The distance of the boat downstream from the bridge when you catch it is______________.

Answer 1

d = 142.5 m

Step-by-step explanation:

This is a vector exercise. Let's calculate how much the boat travels in the 40s

d₀ =
`v_(b)` t

d₀ = 0.75 40

d₀ = 30 m

Let's write the kinematic equations

Boat

x = d₀ +
`v_(b)` t

x = 0 +
`v_(h)` t

At the meeting point the coordinate is the same for both

d₀ +
`v_(b)` t =
`v_(h)` t

t (
`v_(h)` -
`v_(b)`) = d₀

t = d₀ / (
`v_(b)`-
`v_(h)`)

The two go in the same direction therefore the speeds have the same sign

t = 30 / (0.95-0.775)

t = 150 s

The distance traveled by man is

d =
`v_(h)` t

d = 0.95 150

d = 142.5 m