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

Answer:

    d = 142.5 m

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