A swimmer wants to cross a river, from point A to point B. The distance d 1 (fro
ID: 2270797 • Letter: A
Question
A swimmer wants to cross a river, from point A to point B. The distance d1 (from A to C) is 200 m, the distance d2 (from C to B) is 150 m, and the speed vr of the current in the river is 5 km/hour. Suppose that the swimmer makes an angle of ?=45degrees (0.785 radians) with respect to the line from A to C, as indicated in the figure.
A swimmer wants to cross a river, from point A to point B. The distance d1 (from A to C) is 200 m, the distance d2 (from C to B) is 150 m, and the speed vr of the current in the river is 5 km/hour. Suppose that the swimmer makes an angle of ?=45degrees (0.785 radians) with respect to the line from A to C, as indicated in the figure. To swim directly from A to B, what speed us, relative to the water, should the swimmer have? Express the swimmer's speed numerically, to three significant figures, in units of kilometers per hour.Explanation / Answer
The question is asking you to find the magnitude of the velocity vector u_s in the diagram, which is the velocity of the swimmer relative to the current. Let's call that velocity u_s = u_sx x + u_sy y, where u_sx is the swimmer's velocity relative to the current in the x-direction, and u_sy is the swimmer's velocity relative to the current in the y-direction. x and y are unit vectors in the x and y directions, respectively. Let's call the velocity of the swimmer relative to the shore v_s, where v_s = (u_sx + 5) x + u_sy y, where 5 is the speed of the current in km/hr. In other words, the velocity of the swimmer relative to the shore is the vector sum of the velocity of the current relative to the shore (+5 km/hr in the x-direction) and the velocity of the swimmer relative to the current (u_s).
The swimmer will swim straight from A to B when his/her velocity vector relative to the shore is parallel to the vector connecting A to B. This vector can be written as d2 x + d1 y = 150 x + 200 y (where 150 and 200 are in meters, and x and y are again unit vectors). Note that the ratio of the x-component to the y-component of this vector is 150/200 = 0.75. The velocity of the swimmer relative to the shore will be parallel to this vector when its x and y components are in the same proportion; that is, when (v_sx)/(v_sy) = (u_sx + 5)/(us_y) = 0.75. Since the swimmer is swimming in a direction 45 degrees counterclockwise from the line A-C, you have u_sx = -u_sy. Therefore we can write (us_x + 5)/(-us_x) = 0.75 in order for the swimer to swim directly from A to B. Upon rearranging we get 1.75*us_x + 5 = 0, or us_x = -2.86 km/hr. The actual speed of the swimmer relative to the current is |u_s| = sqrt(u_sx^2 + u_sy^2) = sqrt[(-2.86 km/hr)^2 + (+2.86 km/hr)^2] = 4.04 km/hr.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.