Aqua dude can swim at a speed of 2.0m/s in still water. A river is 48 m wide and has a current of 1.5 m/s [E]. He starts at the Southbank and swims so that he is always headed directly across the river.
A) what is Aqua dudes velocity relative to the shore?
B) how long does it take him to cross the river?
I did b since current doesn't affect an object since he's going in the same direction of he would in still water. So:
T=d/v
=48m/2.0m/s
T=24s
The answer is correct I verified it. Could some please explain how I would solve A. The answer given is VAS=2.5m/s[N 37• E]. I just don't understand. Thank you
A. X = 1.5 m/s
Y = 2 m/s.
TanA = Y/X = 2/1.5 = 1.33333
A = 53o, CCW = 90-53 = 37o E. of N. or
[N37E].
To solve part A, we need to find Aqua dude's velocity relative to the shore. This means we want to determine the combination of his swimming speed and the river's current that ultimately affects his movement in relation to the shore.
To analyze this situation, we can break down Aqua dude's velocity into two components: the component parallel to the river's flow (eastward), and the component perpendicular to the river's flow (northward).
The velocity component parallel to the river's flow (eastward) is the velocity of the river current itself, which is given as 1.5 m/s [E].
The velocity component perpendicular to the river's flow (northward) is Aqua dude's swimming speed. Since he is always headed directly across the river, his northward velocity relative to the river's flow would be his swimming speed of 2.0 m/s.
Now, to find Aqua dude's velocity relative to the shore, we can use vector addition. We add these two velocity components together to get the result.
Given that the velocities are at right angles to each other, we can use the Pythagorean theorem to find the magnitude of the resultant velocity:
V^2 = VA^2 + VS^2
Where:
V = resultant velocity relative to the shore (what we're trying to find)
VA = velocity component parallel to the river's flow (1.5 m/s [E])
VS = velocity component perpendicular to the river's flow (2.0 m/s)
Plugging in the values, we get:
V^2 = (1.5 m/s)^2 + (2.0 m/s)^2
V^2 = 2.25 m^2/s^2 + 4.0 m^2/s^2
V^2 = 6.25 m^2/s^2
Taking the square root of both sides, we find:
V = 2.5 m/s
So Aqua dude's velocity relative to the shore is 2.5 m/s.
To determine the direction of the relative velocity, we can use trigonometry. The northward component (2.0 m/s) and the eastward component (1.5 m/s) form a right triangle. The angle θ between Aqua dude's resultant velocity and the north direction (as specified in the answer) can be calculated using the tangent function:
tan(θ) = VA / VS
tan(θ) = (1.5 m/s) / (2.0 m/s)
θ ≈ 37 degrees
Therefore, Aqua dude's velocity relative to the shore is 2.5 m/s [N 37° E].