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Wewaii [24]
3 years ago
9

Please help!!! A river has a constant current of 3 km per hour. If a motorboat, capable of maintaining a constant speed of 20km

per hour, tries to reach a point directly opposite the dock, what will his actual speed and direction be? If the river is 2 km wide, how long will it take to cross?
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
4 0
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have: 

<span>theta = arcsin(3/20) = approx. 8.63° </span>

<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>

<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>

<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
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aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0
3 years ago
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