A swimmer wants to cross a river, from point A to point B. The distance d_1 (from A to C) is 200 m, the distance d_2 (from C to B) is 1500 m, and the speed V_r of the current in the river is 5 km/h. Suppose that the swimmer makes an angle of 45 degrees(0.785 radians) with respect to the line from A to C, as indicated in the figure. Question #1) To swim directly from A to B, what speed U_s, relative to the water, should the swimmer have? Express the swimmer's speed numerically, to three significant figures, in units of kilometers per hour. U_s[/tex] = __________km/h ok here's what i think i should do. first i need to find the distance from A to B. so of course a^2 + b^2 = c^2 1500^2 + 200^2 = C^2 C= 1513.37m from distance A to B ok i know that it makes a 45 degrees angle with respect to the line from A to C. and the inner angle(dont know the math term) or the angle next to it is also 45 degrees(90-45). that means i can find the initial velocity of Vx and Vy. have to convert meters to km first right? so 1500 would be 1.5 Vx = 1.5km*cos(45) = 1.06 km/h Vy = .2km*sin(45) = .1414 km/h ok so i know the distance from A to B, the Vx and Vy initial velocities. im stuck here, dont know what else to do. the picture is attached.
First, you mentioned the distance between B and C as 150 m, and later said it was 1500. A second thing to keep in watch, when you converted m to km, and multiplied it by cos(45) or sin(45), you changed distance to velocity by adding the "/h". Really, 1.5km * cos(45) = 1.06 km Keep in mind that the river velocity is 5 km horizontally. The velocity of the swimmer will need to be opposite of that, as if he or she is to swim in a straight line, the overall horizontal movement of the swimmer will need to be 0. Now, that you know the horizontal velocity, you can find the overall velocity of the swimmer.
i had a typo while changing latex into normal text. it's 1500m from C to B. sorry about that. how would i find the overall velocity of the swimmer? do i need to use the final velocity formula(v(t) = V(0) + at)? if the horizontal velocity is 5km, wouldnt it help the swimmer to get to points A to B faster? im really trying, but i just dont get it.
Heres how I did it. I figure this has been unanswered long enough to give more than a hint. I hope im right. Well start out in vectors. Assume to the left is positive, and up is positive. Velocity of Swimmer (relative to standing water) = Vx i + Vy j We know that since its at a 45 degree, Vx/Vy = 1, let Vx=Vy=Vs So Velocity of Swimmer (relative to standing water) = Vs i + Vs j The total velocity of the Swimmer as he is crossing the river(relative to the shore) is Vs + Vw = Vs i + Vs j - 5000m/hr i = (Vs - 5000) i + Vs j Vw is the velocity of the water. Now we just use the simple : x=vt x=Vx*t y=Vy*t for x : -150 = (Vs-5000)* t /// REMEMBER he moves in the negative direction. for y : 200 = Vs * t Solve for t = 200/Vs Plug into first eq: -150 = (Vs-5000)*(200/Vs) Solve for Vs Vs = 2857.14 m/hour == 2.86 km/hr