A student swims at 4m/s. She is swimming from west to east across a river that's is flowing 2.2m/s [S].
A) if the river is 100 m wide, what is the quickest time that she can swim the entire width?
B) how far downstream will she be?
C) if she decides to go directly across the river, what direction should she swim?
D) how long will it take her to cross the river going this direction?
quiz tomorrow on these type of questions I need help!!!!
a. d = V*t = 100 m.
4*t = 100.
t =25 s.
b. Tan A = Y/X = -2.2/4 = -0.550.
A = -28.8o = 28.8o S. of E.
Tan 28.8 = Y/X = d/100
d = 100*Tan28.8 = 55 m. downstream.
c. Direction = 28.8o N. of E. to offset
the current.
To solve this problem, we'll use vector addition. Let's break down the given information:
The speed of the student relative to the ground (also called the resultant velocity) is 4 m/s from the west to east direction.
The river is flowing from the south at a speed of 2.2 m/s.
A) The quickest time she can swim the entire width of the river can be found by calculating the resultant velocity. We'll use the Pythagorean theorem for vector addition.
The resultant velocity vector is obtained by taking the square root of the sum of the squares of the original vectors:
Resultant Velocity = √(4^2 + 2.2^2) = √(16 + 4.84) = √20.84 ≈ 4.57 m/s
Now, we can find the time it takes to cross the river using the formula: time = distance/speed.
Distance = 100 m (given), Speed = 4.57 m/s (resultant velocity)
Time to cross the river = 100 m / 4.57 m/s ≈ 21.87 seconds
Therefore, the quickest time she can swim the entire width is approximately 21.87 seconds.
B) To determine how far downstream she will be from her starting position, we'll need to calculate the horizontal component of the river's velocity.
The horizontal component of the river's velocity is the vector in the east to west direction, and it can be found using the formula:
Horizontal Component = River Velocity * cos(θ)
Considering the river flows south, we need to find the angle that the resultant velocity makes with the east direction. We can use the inverse tangent (arctan) of the vertical component of the river's velocity divided by the horizontal component of the river's velocity:
θ = arctan(2.2 m/s / 4 m/s) ≈ arctan(0.55) ≈ 29.74 degrees
Now, we can calculate the horizontal component of the river's velocity:
Horizontal Component = 2.2 m/s * cos(29.74 degrees)
Therefore, the student will be downstream by:
Distance downstream = Horizontal Component * Time to cross the river.
Substituting the values:
Distance downstream = (2.2 m/s * cos(29.74 degrees)) * 21.87 seconds
C) If the student decides to go directly across the river, she should swim perpendicular to the direction of the resultant velocity. This would be an angle of 90 degrees from west to east.
D) To find out how long it will take her to cross the river in this direction, we can calculate the time using the formula: time = distance/speed.
We know the distance is 100 m (width of the river), and the student's speed relative to the ground is 4 m/s.
Time to cross directly = 100 m / 4 m/s = 25 seconds.
Therefore, it will take her approximately 25 seconds to cross the river in a straight line.