A boat is steered at angle ? = 30.0 degrees west of north with a speed v bw rela

Question

A boat is steered at angle ? = 30.0 degrees west of north with a speed vbwrelative to the water. The water due east with a speed vws = 0.40 km/hr with respect to the shore. A person on the shore sees the boat traveling directly northward across the river.  It take the boat 29.0 min to cross the river.

a) What is the x component (EW direction) of the velocity of the boat relative to the shore?

b) What is the x component (EW direction) of the velocity of the boat relative to the water?

c) What is the y component (NS direction) of the velocity of the boat relative to the shore?

d) What is, in km, the width of the river at that location?

Explanation / Answer

a.) x-component is 0km/hr.

Boat is moving directly northward relative to someone on the shore, so it obviously does not have an x-component of speed when being measured from the shore.

b.) Boat must be moving 0.4km/hr WEST

We know that the boat's x-component of velocity and the water's x-component of velocity must cancel, thus the boat's x-component of velocity must be equal and opposite to the water's x-component when measured relative to the water itself. Since the water is moving 0.4km/hr EAST, the boat must be moving 0.4km/hr WEST relative to the water.

c.)what the boat's overall velocity is given that it is pointed at 30 degrees west of north and that its x-component is 0.4km/hr west. We could use Sin(a) or Cos(a) to solve for the absolute magnitude of its velocity, but this isn't what the question asks, it wants the y-component. Furthermore, no other part of this question needs the absolute magnitude of the boat's velocity relative to the water, so we'll just use Tan(a) instead to solve for the y-component immediately. Given that Tan(a) = opp/adj, we just need to determine which is the opposite and which is the adjacent, carefully noting that this is 30 degrees measured from DUE NORTH. Imagine the three velocity components of the boat relative to the water (the x-component, the y-component, and absolute magnitude) as being three sides to a right trangle, where the x-component is the bottom along the negative x-axis, the y-component goes straight up from the end of the x-component at a right angle, and the absolute magnitude goes from the origin to the end of the y-component vector at an angle 30 degrees counter-clockwise from the positive y-axis. The angle between the absolute magnitude vector and the x-component vector is actually 90deg - 30deg = 60deg. We can use this angle with Tan(a) to determine the value of the y-component given that the y-component is the opposite side of this angle and the x-component is the adjacent. Thus Tan(60deg) = y-component / 0.4km/hr, or y-component = 0.4km/hr*Tan(60deg) = 0.6928km/hr. Similarly, we can use some geometric trickery to say Tan(30deg) = 0.4km/hr / y-component --> y-component = 0.4km/hr / Tan(30deg) = 0.6928km/hr. If we had known the absolute magnitude of the velocity outright, we could have also used the Pythagorean Theorem to determine the y-component of velocity.

d) Given that we now know that the boat is going to reach the other side of the river in 29 minutes, we use v = d/t to determine the distance d in km that the boat travels in 29 minutes, which gives us the width. Since v =0.6928km/hr in the direction across the river, and t = 29mins/60mins = 0.4833 hr, we can say that d = v*t = 0.6928km/hr*0.4833hr = 0.33485km.

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