A man can row a boat at 13m/s in still water. If the aims at crossing to the opposite bank of a river flowing 5m/s, at what angle to the bank of the river must he row the boat?
Vr = Vb - 5i = 13 = Resultant velocity.
Vb = 13 + 5i = 13.9 m/s[21o]. = Velocity of the boat.
Direction = 21deg. N. of E.
draw a picture
the cosine of the angle is ... 5 / 13
To determine the angle at which the man needs to row the boat, we can use the concept of vector addition. The velocity of the boat in still water can be represented as VB, and the velocity of the river flow can be represented as VR.
The resultant velocity, let's call it VT (the velocity of the boat relative to the ground), is the vector sum of VB and VR.
Using the Pythagorean theorem, we can calculate the magnitude of VT:
VT^2 = VB^2 + VR^2
Substituting the known values:
VT^2 = (13 m/s)^2 + (5 m/s)^2
VT^2 = 169 m^2/s^2 + 25 m^2/s^2
VT^2 = 194 m^2/s^2
Taking the square root of both sides to find VT:
VT ≈ √194 ≈ 13.93 m/s
Now, we can find the angle θ that the boat must be rowed relative to the riverbank using trigonometry.
Tan(θ) = VR / VB
Tan(θ) = 5 m/s / 13 m/s
Tan(θ) ≈ 0.38
θ ≈ Tan^−1(0.38)
Using inverse tangent function to find θ:
θ ≈ 21.8 degrees (rounded to one decimal place)
Therefore, the man must row the boat at an angle of approximately 21.8 degrees to the bank of the river in order to cross it directly.
To find the angle at which the man must row the boat, we need to consider the velocity vectors of the boat and the river. Let's break it down step by step:
1. Draw a diagram: Draw a diagram representing the boat's velocity (13 m/s) and the river's velocity (5 m/s). Label the angle between them as θ.
________ River velocity (5 m/s)
|
|\
| \
| \
| \
| \
| \
|θ Boat velocity (13 m/s)
|
----------------------->
River
2. Break down the velocities: We can break down the velocities into their horizontal (x-axis) and vertical (y-axis) components. The boat's horizontal component will be its speed in still water, which is 13 m/s. The vertical component is zero since the boat does not move up or down. The river's horizontal component is 5 m/s, and its vertical component is also zero.
________ (5 m/s, 0 m/s)
|
|\
| \
| \
| \
| \
| \
|θ 13 m/s, 0 m/s
|
----------------------->
River
3. Add up the velocities: Add the horizontal components and vertical components separately to get the resulting velocity vector. The resulting vector will be the actual velocity of the boat relative to the river.
________ (5 m/s, 0 m/s)
|\ |
| \ |
| \ |
| \ |
| \ |
| \ |
|θ 13 m/s, 0 m/s
|
----------------------->
River
________ (5 m/s, 0 m/s)
|\ |
| \ |
| \ |
| \ |
| \ |
| \ |
|θ 13 m/s*cos(θ), 13 m/s*sin(θ)
|
----------------------->
Resultant (Boat + River)
4. Solve for θ: Since we now have the x and y components of the resultant velocity vector, we can use the trigonometric relationship:
tan(θ) = (13 m/s * sin(θ)) / (5 m/s + 13 m/s * cos(θ))
To solve for θ, we need to rearrange the equation and isolate θ on one side. However, this equation does not have a simple algebraic solution. It requires numerical methods or a calculator to find the angle θ.
So, to find the angle θ, plug the equation into a calculator or use a computer program that can solve trigonometric equations numerically.