a boat can sail at 27 m.s in still water. the boat must cross a river that flows at 9 m.s towards the east . if the boat crosses the river by going up stream at an angle of 23 south, what will the size and angle of the resulting velocity be?
23o W. of S. = 247o CCW.
Vr = 27m/s[247o] + 9m/s.
Vr = 27*Cos247 + 27*sin247 + 9 =
-10.55 - 24.85i + 9 = -1.55 - 24.85i =
24.9m/s[86.4o] S. of W.
To find the resulting velocity of the boat, we can use vector addition. Let's break down the velocities into their horizontal (x-axis) and vertical (y-axis) components.
Given:
Velocity of the boat in still water, v = 27 m/s
Velocity of the river, vr = 9 m/s
Horizontal Component:
The boat's horizontal component of velocity will remain unaffected by the river flow since the river flows towards the east, and the boat is crossing at an angle.
Vertical Component:
The boat's vertical component of velocity will be affected by both its own speed and the river's flow. The vertical component of the boat's velocity is given by the equation:
vb = v * sin(angle)
where vb is the vertical component of the boat's velocity and angle is the angle at which it is crossing the river.
vb = 27 * sin(23°)
vb ≈ 27 * 0.3907
vb ≈ 10.5519 m/s
Now, let's find the horizontal and vertical components of the river's velocity.
Horizontal Component of River's Velocity:
The river's horizontal component of velocity will be vr since it flows towards the east.
Vertical Component of River's Velocity:
The river's vertical component of velocity (vr) will be unaffected as the boat is crossing at an angle.
Now let's find the resultant velocity by adding the horizontal and vertical components separately.
Horizontal Component of Resultant Velocity:
The horizontal component of resultant velocity (res_vx) is the sum of the boat's horizontal velocity component and the river's horizontal velocity component.
res_vx = v + vr
res_vx = 27 + 9
res_vx = 36 m/s
Vertical Component of Resultant Velocity:
The vertical component of resultant velocity (res_vy) is the difference between the boat's vertical velocity component and the river's vertical velocity component.
res_vy = vb - 0
res_vy = vb
res_vy ≈ 10.5519 m/s
To find the magnitude and angle of the resulting velocity, we can use the Pythagorean theorem and trigonometry.
Magnitude of Resultant Velocity:
res_v = sqrt(res_vx^2 + res_vy^2)
res_v = sqrt((36)^2 + (10.5519)^2)
res_v ≈ sqrt(1296 + 111.305)
res_v ≈ sqrt(1407.305)
res_v ≈ 37.51 m/s
Angle of Resultant Velocity:
tan(angle) = res_vy / res_vx
tan(angle) = 10.5519 / 36
angle = arctan(10.5519/36)
angle ≈ 16.75°
Therefore, the magnitude of the resulting velocity is approximately 37.51 m/s, and the angle of the resulting velocity is approximately 16.75° with respect to the horizontal axis.