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?
Are you trying to go straight across the river by aiming upstream? What is meant by an angle of 23 south? South of what?
23 degrees south
To find the size and angle of the resulting velocity, we need to break down the velocities into their components, both horizontally and vertically.
First, let's consider the boat's velocity in still water. It is given as 27 m/s, and since there is no current, this velocity only has a horizontal component. Let's call it Vb (velocity of the boat in still water).
Vb (horizontal component) = 27 m/s
Vb (vertical component) = 0 m/s (since there is no current)
Next, let's consider the velocity of the river's flow. It is given as 9 m/s towards the east, so it only has a horizontal component. Let's call it Vr (velocity of the river's flow).
Vr (horizontal component) = 9 m/s
Vr (vertical component) = 0 m/s (as there is no vertical flow mentioned)
Now, we can find the resulting velocity of the boat by considering the boat's velocity and the river's velocity as vectors. We can represent the velocity vectors graphically using a triangle or mathematically using vector addition.
To find the horizontal component of the resulting velocity, we need to subtract the horizontal component of the river's velocity from the horizontal component of the boat's velocity:
Vres (horizontal component) = Vb (horizontal component) - Vr (horizontal component)
= 27 m/s - 9 m/s
= 18 m/s
To find the vertical component of the resulting velocity, we need to add the vertical components of both velocities:
Vres (vertical component) = Vb (vertical component) + Vr (vertical component)
= 0 m/s + 0 m/s
= 0 m/s
Thus, the resulting velocity has a horizontal component of 18 m/s and a vertical component of 0 m/s.
Now, let's find the size and angle of the resulting velocity using trigonometry.
Size of the resulting velocity (Vres) = √[(Vres horizontal component)^2 + (Vres vertical component)^2]
= √[(18 m/s)^2 + (0 m/s)^2]
= √(324 m^2/s^2)
≈ 18 m/s
The angle of the resulting velocity (θ) can be found using the inverse tangent (arctan) function:
θ = arctan(Vres vertical component / Vres horizontal component)
= arctan(0 m/s / 18 m/s)
= arctan(0)
= 0°
Therefore, the size of the resulting velocity is approximately 18 m/s, and the angle of the resulting velocity is 0° (horizontal).