A boy can swim with a speed of 26metre per minute in still water. Be wants to swim across a 150metere wide river from point A to point B directly opposite on the other side of the River.The river flows at 10metre per minute (a)if he always swims in the direction parallel toAB,find how far he lands downstream of B. (b) in what direction relative to the bankmust the boy swim so as to cross directly from A to B
1)10/26=X/150 X=1500/26=57.7m 2)costheta=10/26 theta=67.3degrees
a. Tan A = Y/X = 10/26 = 0.38462
A = 21.04o E of N.
Tan A = d/150
d = 150*Tan 21.04 = 57.7 m = Distance
downstream.
b. To offset the affect of the current,
the swimmer must head 21.04o W. of N.
Vsc = Vs + Vc = 26i
Vs + 10 = 26i
Vs = -10 + 26i
Tan A = 26/-10 = -2.600
X = -68.96 = 68.96o CW from -x axis
180-68.96 = 111.04o CCW. = 21.04o W. of
N.
To solve this problem, we need to break it down into two components: the horizontal component of the boy's motion and the vertical component due to the river flow.
(a) To find how far the boy lands downstream of point B, we need to calculate the time it takes for him to swim across the river. We can use the formula: time = distance / speed.
The distance the boy needs to swim across the river is 150 meters (the width of the river), and his swimming speed is 26 meters per minute. Therefore, the time it takes for him to swim across the river is: time = 150 / 26.
Now, we need to calculate the distance the river flow has carried the boy downstream during this time. The river flows at a speed of 10 meters per minute. So, the distance the boy moves downstream due to the river flow is: distance = time × flow rate = (150 / 26) × 10.
Therefore, the boy lands downstream from point B by a distance of (150 / 26) × 10 meters.
(b) To determine the direction relative to the bank that the boy must swim in order to cross directly from point A to point B, we need to take into account both the direction of the river flow and the direction the boy wants to reach.
Since the river flows parallel to the line AB, the boy needs to swim at an angle to counteract the river flow and reach point B directly. This means the angle between his swimming direction and the line AB should be opposite to the angle between the direction of the river flow and the line AB.
Essentially, the boy should swim at an angle so that the resultant vector (the combination of his swimming speed and the river flow) forms a straight line across the river from point A to point B.
To determine the specific angle, you can use trigonometry. The tangent of the angle can be calculated as the ratio of the vertical component of the river flow to the horizontal component of the boy's velocity.
tan(θ) = river flow / swimming speed
Using the given values, tan(θ) = 10 / 26.
Finally, solve for θ by taking the arctan of both sides: θ = arctan(10 / 26).
Now you know the angle relative to the bank that the boy must swim in order to cross directly from point A to point B.