|Sep13-04, 03:47 PM||#1|
Crossing a River
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.
|Sep13-04, 08:29 PM||#2|
k changed latex into normal text, i guess latex is not working right now.
|Sep13-04, 10:08 PM||#3|
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.
|Sep13-04, 11:09 PM||#4|
Crossing a River
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.
|Sep14-04, 05:15 PM||#5|
Has anyone had any luck with this problem?
|Sep14-04, 06:14 PM||#6|
oops, my bad agian, it's actually 150m from C to B lol. sorry...
|Sep14-04, 06:53 PM||#7|
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
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
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
|Sep16-04, 10:09 AM||#8|
Hi everyone. Just figured out the answer to be 4.04km/h.
|Sep16-04, 06:24 PM||#9|
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