How Do Canoeists A and B Coordinate Their Arrival Using Relative Motion?

[grace lotz]

please help with relative motion problem!

I need some help with this question, i know the axis needs to be rotated but I'm not sure how to go about doing that. I also need to know how to do it using the component method.

-two canoeists, A and B, live on opposite shores of a 300m wide river that flows east at .80m/s. A lives on the north shore and B lives on the south shore. They both set out to visit a mutual friend X who lives on the north shore at a point 200m upstream from A and 200m downstream for B. Both canoeists can propel their canoes at 2.4m/s through the water. How much time must canoeist A wait after canoeist B sets out so that they both arrive at X at the same time? Both canoeists make their respective trips by the most direct routes.

THANKYOU!

 

[Andrew Mason]

I need some help with this question, i know the axis needs to be rotated but I'm not sure how to go about doing that. I also need to know how to do it using the component method.

-two canoeists, A and B, live on opposite shores of a 300m wide river that flows east at .80m/s. A lives on the north shore and B lives on the south shore. They both set out to visit a mutual friend X who lives on the north shore at a point 200m upstream from A and 200m downstream for B. Both canoeists can propel their canoes at 2.4m/s through the water. How much time must canoeist A wait after canoeist B sets out so that they both arrive at X at the same time? Both canoeists make their respective trips by the most direct routes.

THANKYOU!

Think in velocity vectors. For B to get to X he moves directly toward X at a velocity consisting the sum of two velocity vectors. You know the magnitude and direction of the current flow vector (.8 m/s east). You know the magnitude of the canoe vector (2.4 m/s). You know the direction of the resultant (from B to X). So the canoe must move with velocity such that its eastward speed / northward speed = 2/3.

 

[wais]

To solve this problem, we can use the relative motion equation: Vab = Vao + Vob. In this equation, Vab represents the velocity of A with respect to B, Vao represents the velocity of A with respect to the observer on shore, and Vob represents the velocity of B with respect to the observer on shore.

First, we need to determine the velocity of A with respect to the observer on shore. Since the river is flowing east at 0.80m/s, A's velocity with respect to the observer on shore will be 2.4m/s + 0.80m/s = 3.2m/s to the east.

Next, we need to determine the velocity of B with respect to the observer on shore. B's velocity with respect to the observer will be 2.4m/s to the east, since they are both traveling in the same direction.

Now, we can plug these values into the relative motion equation: Vab = 3.2m/s + 2.4m/s = 5.6m/s. This means that A is moving 5.6m/s faster than B.

To find the time that A must wait after B sets out, we need to use the distance formula: d = rt. Since both A and B are traveling the same distance (200m), we can set up the following equation:

200m = 5.6m/s * t

Solving for t, we get t = 35.71 seconds. This means that A must wait 35.71 seconds after B sets out in order to arrive at X at the same time.

To solve this using the component method, we can break down the velocities into their x and y components. The velocity of A with respect to the observer on shore will have a y component of 0m/s and an x component of 3.2m/s. The velocity of B with respect to the observer on shore will have both x and y components of 2.4m/s.

Using the Pythagorean theorem, we can find the magnitude of the relative velocity between A and B: Vab = √(3.2^2 + 2.4^2) = 3.92m/s.

Now, we can use the distance formula again to find the time that A must wait after B sets out:

200m