Respuesta :
Answer:
Explanation:
Given
Width of river=200 m
Speed of river[tex](v_r)[/tex]=2 m/s
Speed of Boat relative to river[tex](v_{br})[/tex]=8 m/s
Boat leaves the bank [tex]30^{\circ}[/tex] west of north
[tex]v_{br}=-8cos30\hat{i}+8sin30\hat{j}[/tex]
[tex]v_{r}=2\hat{i}[/tex]
Therefore velocity of boat w.r.t ground
[tex]v_b=v_{br}+v_r=-8cos30\hat{i}+8sin30\hat{j}+2\hat{i}[/tex]
[tex]v_b=-4.928\hat{i}+4\hat{j}[/tex]
Magnitude of boat speed[tex]=\sqrt{4.928^2+4^2}=6.34 m/s[/tex]
For direction
[tex]tan\theta =\frac{4}{4.928}[/tex]
[tex]\theta =39.06^{\circ}[/tex] w.r.t to west
to cross the river its north component will help so
time taken [tex]t=\frac{200}{4}=50 s[/tex]
The resultant velocity is the vector sum of the velocity of the boat and
the velocity of the river.
Correct responses:
The (approximate) values are;
(a) 7.2 m/s
(b) 106.1°
(c) 28.9 s
Methods used to calculate magnitude and direction of aa vector
Width of the river = 200 m
The direction the river flows = Due east
The speed of the river = 2.0 m/s
Speed of the boat = 8.0 m/s
Direction of the boat = 30° west of north
(a) The magnitude of the resultant velocity of the boat is found as follows;
Vector form of the velocities.
[tex]\vec v_{river}[/tex] = 2.0·i
[tex]\vec {v}_{boat}[/tex] = -8.0·sin(30°)·i + 8.0·cos(30°)·j = -4.0·i + 4.0·√3·j
Therefore;
[tex]\vec v_{resultant}[/tex] = 2.0·i - 4.0·i + 4.0·√3·j = -2.0·i + 4.0·√3·j
[tex]| \vec {v}_{resultant}|[/tex] = v = [tex]\sqrt{(-2)^2 + \left(4.0\cdot \sqrt{3} \right) } = \mathbf{ 2\cdot \sqrt{13}}[/tex]
- The magnitude of the (resultant) velocity of the boat, v = [tex]2 \cdot \sqrt{13}[/tex] m/s ≈ 7.2 m/s
(b) Let θ represent the direction of the boat's relative velocity to the ground, we have;
[tex]tan\left(\theta \right) = \mathbf{\dfrac{4.0 \cdot \sqrt{3} }{-2.0}}[/tex]
Therefore;
[tex]\theta = arctan \left(\dfrac{4.0 \cdot \sqrt{3} }{-2.0} \right) \approx \mathbf{ -73.9^{\circ}}[/tex]
Which gives;
- The direction angle relative to the positive x-axis, ∅ ≈ 180° - 73.9° = 106.1°
(c) The time it takes the boat to cross the river is given as follows;
Direction across the river = Due north
Distance across the river = 200 m
Component of the velocity of the boat in the north direction = 4.0·√3 m/s
[tex]Time = \mathbf{ \dfrac{Distance}{Velocity}}[/tex]
Time it takes the boat to cross the river is therefore;
- [tex]Time = \dfrac{200 \, m}{4.0 \cdot \sqrt{3} \, m/s} = \dfrac{50 \cdot \sqrt{3} }{3} \, s \approx \underline{28.9 \, s}[/tex]
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