A 130-m-wide river flows due east at a uniform speed of 4.1m/s. A boat with a speed of 6.6m/s relative to the water leaves the south bank pointed in a direction 29o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground? Give the direction as the angle of the velocity from due north, positive if to the east and negative if to the west. (c) How long does it take for the boat to cross the river?
a,b. Vb = 6.6m/s[119o] + 4.1 =
-3.2 + 5.77i + 4.1 = 0.9 + 5.77i =
5.84m/s[81o] N. of E. = 9o E. of N.
c. 130m * 1s/5.77m = 22.53 s.
To find the boat's velocity relative to the ground, we need to consider both the velocity of the boat relative to the water and the velocity of the water relative to the ground.
Let's break down the problem step by step:
(a) First, let's find the magnitude of the boat's velocity relative to the ground using vector addition. We can use the Pythagorean theorem to find the magnitude of the resultant vector. Since the boat's velocity is given in terms of its components, we can use trigonometric functions to find the horizontal and vertical components of the boat's velocity relative to the ground.
The horizontal component of the boat's velocity relative to the ground is given by:
Vx = (velocity of the boat relative to water) * cos(angle between boat's velocity and the horizontal)
Vx = 6.6 m/s * cos(29°)
The vertical component of the boat's velocity relative to the ground is given by:
Vy = (velocity of the boat relative to water) * sin(angle between boat's velocity and the horizontal)
Vy = 6.6 m/s * sin(29°)
Using these components, we can find the magnitude of the boat's velocity relative to the ground using the Pythagorean theorem:
V = √(Vx^2 + Vy^2)
(b) To find the direction of the boat's velocity relative to the ground, we can use trigonometric functions again. The angle can be found using the inverse tangent function:
θ = atan(Vy / Vx)
(c) To find the time it takes for the boat to cross the river, we need to consider the distance and the horizontal component of the boat's velocity relative to the ground. The distance across the river (from the south bank to the north bank) is given as 130 meters, and the horizontal component of the boat's velocity relative to the ground (Vx) remains constant during the crossing. Using the formula:
Time = Distance / Vx
Now let's calculate the values:
(a) Magnitude of the boat's velocity relative to the ground:
Vx = 6.6 m/s * cos(29°) ≈ 5.828 m/s
Vy = 6.6 m/s * sin(29°) ≈ 3.065 m/s
V = √(5.828^2 + 3.065^2) ≈ 6.585 m/s
Therefore, the magnitude of the boat's velocity relative to the ground is approximately 6.585 m/s.
(b) Direction of the boat's velocity relative to the ground:
θ = atan(3.065 m/s / 5.828 m/s)
θ ≈ 28.5°
The direction of the boat's velocity relative to the ground is approximately 28.5° east of due north.
(c) Time it takes for the boat to cross the river:
Time = 130 m / 5.828 m/s ≈ 22.34 seconds
Therefore, it takes approximately 22.34 seconds for the boat to cross the river.