A swimmer heads directly across a river, swimming at 1.10 m/s relative to the water. She arrives at a point 60.0 m downstream from the point directly across the river, 80.0 m wide. What is the speed of the river current?
What is the swimmer's speed relative to the shore?
In what direction (as an angle relative to a direct line across the river) should the swimmer aim instead, so that she arrives at the point directly opposite her starting point?
Elena, i have a little problem with your calculation. I was thinking the velocity of the swimmer should tally with one of the adjacent sides of the triangle and not the hypothenus. the hypothenus should be in line with the resultant velocity of the swimmer.
(a)
s=sqrt(80²+60²) = 100 m.
sinα=60/100=v/1.1.
v=60•1.1/100 = 0.66 m/s.
(b)
sin φ= 0.66/1.1 = 0.06
φ=arcsin 0.06 = 3.44º
Why did the swimmer go to math class? Because she wanted to solve the current-ly perplexing river problem!
To find the speed of the river current, let's assume it as 'v'. Now, the swimmer is swimming at 1.10 m/s relative to the water, and she arrives at a point 60.0 m downstream from the point directly across the river. This means that the time taken to cross the river is the same as the time taken to be carried downstream. So we can say:
Time taken to cross the river = Time taken to be carried downstream.
Now, the time taken to cross the river is given by (distance / speed). The distance she needs to cross is the width of the river, which is 80.0 m. Her swimming speed relative to the water is 1.10 m/s. Therefore:
80.0 m / 1.10 m/s = 72.73 s
So, the time taken for her to be carried downstream is also 72.73 s. The distance she is carried downstream is given as 60.0 m.
Now, we can find the speed of the river current by using the formula (distance / time). The distance is 60.0 m and the time is 72.73 s:
Speed of the river current = 60.0 m / 72.73 s ≈ 0.82 m/s
To find the swimmer's speed relative to the shore, we can use the Pythagorean theorem. The swimmer's speed relative to the water is 1.10 m/s, and the speed of the river current is 0.82 m/s.
Using the Pythagorean theorem:
Swimmer's speed relative to the shore = √(1.10^2 - 0.82^2) ≈ 0.67 m/s
Now, to find the direction the swimmer should aim instead to arrive at the point directly opposite her starting point, we need to consider the velocity vectors. Since the swimmer needs to counteract the current, she should aim slightly upstream.
The angle can be found using trigonometry. The tangent of the angle is given by (speed of the current / speed of the swimmer relative to the shore). Let's plug in our values:
Tangent of the angle = 0.82 m/s / 0.67 m/s
Using inverse tangent, we can find the angle:
Angle ≈ Inverse tangent (0.82/0.67) ≈ 51.8°
So, the swimmer should aim at an angle of approximately 51.8° upstream to reach the point directly opposite her starting point.
Now, wasn't that a stream of mathematical humor?
To find the speed of the river current, we can use the formula:
v = d / t
where v is the speed, d is the distance, and t is the time taken to cover that distance.
In this case, we know that the swimmer crossed a distance of 80.0 m downstream (since the river is 80.0 m wide) and arrived at a point 60.0 m downstream from the point directly across the river. This means that the swimmer was pushed 60.0 m downstream while swimming across the river.
We can assume that the time taken to swim across the river is the same as the time taken to be pushed downstream. Therefore, the time is the same for both distances.
Using the equation v = d / t and rearranging it to t = d / v, we can calculate the time taken to cover each distance.
Time taken to cover the distance across the river (d1):
t1 = 80.0 m / 1.10 m/s = 72.73 s
Time taken to cover the distance downstream (d2):
t2 = 60.0 m / v (since we don't know the current speed yet)
Since t1 = t2, we can set up an equation:
80.0 m / 1.10 m/s = 60.0 m / v
Solving for v, we get:
v = (60.0 m / 80.0 m) * 1.10 m/s = 0.825 m/s
Therefore, the speed of the river current is 0.825 m/s.
To find the swimmer's speed relative to the shore, we can use the Pythagorean theorem. Since the swimmer is moving both horizontally (across the river) and vertically (due to the river current), the swimmer's speed relative to the shore can be considered as the hypotenuse of a right triangle, with the swimmer's speed across the river as one side and the speed of the river current as the other side.
Using the Pythagorean theorem:
Speed relative to the shore = sqrt((1.10 m/s)^2 + (0.825 m/s)^2)
= sqrt(1.21 m^2/s^2 + 0.680625 m^2/s^2)
= sqrt(1.890625 m^2/s^2)
= 1.376 m/s
Therefore, the swimmer's speed relative to the shore is approximately 1.376 m/s.
To determine the direction that the swimmer should aim for in order to arrive at the point directly opposite her starting point, we need to consider the angle relative to a direct line across the river.
Since the river current pushes the swimmer downstream, the swimmer needs to aim more upstream than directly across the river. To calculate this angle, we can use trigonometry.
Let's call the angle θ.
tan(θ) = opposite / adjacent
tan(θ) = (speed of the river current) / (speed across the river)
Plugging in the values we have:
tan(θ) = 0.825 m/s / 1.10 m/s
tan(θ) = 0.75
Taking the inverse tangent (arctan) of both sides to solve for θ:
θ = arctan(0.75)
θ ≈ 36.9 degrees
Therefore, the swimmer should aim at approximately 36.9 degrees (or upstream) from the direct line across the river in order to arrive at the point directly opposite her starting point.