a river flows due south with a speed of 2.0m/s. a man steers a motorboat across the river;his velosity is 4.2m/s due east.the river is 800m wide.
(A) what is his velocity(magnitude and direction)relative to the earth?
(B)how much time is required to cross the river?
b. time to go across: distanceEast/VelocityEast
a. he is heading upstream:
direction=arctan2/4.2
magnitude: speed=sqrt(4.2^2+2^2)
To solve this problem, we will use vector addition to find the man's velocity relative to the earth.
(A) To find the magnitude of the velocity relative to the earth, we can use the Pythagorean theorem. The magnitude of the velocity relative to the earth (Vearth) is given by:
Vearth = √(Vriver^2 + Vboat^2)
where Vriver is the velocity of the river, and Vboat is the velocity of the boat.
In this case, the velocity of the river (Vriver) is given as 2.0 m/s, and the velocity of the boat (Vboat) is given as 4.2 m/s. Plugging these values into the formula, we get:
Vearth = √(2.0^2 + 4.2^2)
= √(4 + 17.64)
= √21.64
= 4.65 m/s
Therefore, the magnitude of the man's velocity relative to the earth is 4.65 m/s.
To find the direction of the velocity relative to the earth, we can use trigonometry. Since the boat is moving due east and the river is flowing due south, the angle between their velocities is 90 degrees. Therefore, the direction of the man's velocity relative to the earth is southeast.
(B) To determine the time required to cross the river, we can use the equation:
time = distance / velocity
In this case, the distance to be crossed is given as 800 meters, and the velocity of the boat (Vboat) is given as 4.2 m/s. Plugging these values into the formula, we get:
time = 800 m / 4.2 m/s
= 190.5 seconds (rounded to one decimal place)
Therefore, it will take approximately 190.5 seconds for the boat to cross the river.