Swimming at 3.00 m/s relative to still water, a swimmer heads directly across a river of width 80 m. He arrives 50 m downstream from the point directly across the river from his starting point.
(a) What is the speed of the current in the river?
(b) At what angle relative to the line perpendicular to the shore should he head to arrive at a point directly opposite his starting point?
time to swim across: 80/3 seconds
water velocity=50/(80/3) m/s
head upstream arc tan 50/80
To determine the speed of the current in the river, we can use the information provided in the question.
Let's define the following variables:
- v is the speed of the swimmer (swimming speed relative to still water)
- c is the speed of the current
- d is the width of the river (80 m)
- x is the distance the swimmer traveled downstream (50 m)
We know that the swimmer's velocity relative to the ground is the vector sum of their swimming velocity and the current velocity. The magnitude of the ground velocity is given by:
v_ground = sqrt(v^2 + c^2)
Since the swimmer is swimming directly across the river, the component of the ground velocity perpendicular to the current is equal to the swimming speed v:
v_perpendicular = v
The component of the ground velocity parallel to the current is equal to the speed of the current:
v_parallel = c
From the given information, we can set up the following equations:
Equation 1: v_perpendicular = v
Equation 2: v_parallel = c
Equation 3: d = v_ground * t (where t is the time it takes to cross the river)
To solve for the speed of the current (c), we need to eliminate t. We can do this by dividing Equation 3 by Equation 1:
d / v_perpendicular = v_ground * t / v
d / v_perpendicular = (v^2 + c^2) / v
d * v = v^2 + c^2
We also know that the swimmer arrived 50 m downstream from the point directly across the river from the starting point. This implies that the time taken to cross the river is equal to the time taken to travel downstream. Using this information, we can set up another equation:
Equation 4: x = v_parallel * t
We can substitute Equation 2 into Equation 4:
x = c * t
Now, we can eliminate t by dividing Equation 4 by Equation 2:
x / c = t
Substituting this value of t into Equation 3:
d = v_ground * (x / c)
Rearranging the equation:
c = (v_ground * x) / d
Finally, we can substitute the expression for the ground velocity:
c = (sqrt(v^2 + c^2) * x) / d
This is a quadratic equation with c appearing on both sides of the equation. To solve for c, we will need to square both sides, rearrange the equation, and solve for c.
Now that we have determined the speed of the current (c), we can move on to part (b) to find the angle relative to the line perpendicular to the shore at which the swimmer should head.
Since the swimmer wants to arrive directly opposite their starting point, they need to account for the current. The angle the swimmer should head will be the angle between the line perpendicular to the shore and the velocity vector of the swimmer relative to the ground.
To find this angle, we can use trigonometry. Let's call this angle θ.
Using the definitions from before:
Opposite side = c (speed of the current)
Adjacent side = v (swimmer's speed relative to still water)
We can use the tangent function to find the angle θ:
tan(θ) = opposite / adjacent
tan(θ) = c / v
Taking the inverse tangent of both sides:
θ = arctan(c / v)
This is the angle at which the swimmer should head relative to the line perpendicular to the shore to arrive directly opposite their starting point.