Question

A 200-m-wide river flows due east at a uniform speed of 2.0 m/s. A boat with a speed of 8.0 m/s relative to the water leaves the south bank pointed in a direction 30° west of north. What are the (a) magnitude and (b) direction of the boat’s velocity relative to the ground? (c) How long does the boat take to cross the river?

Answer 1

Answer:

Explanation:

Given

Width of river=200 m

Speed of river$(v_r)$=2 m/s

Speed of Boat relative to river$(v_{br})$=8 m/s

Boat leaves the bank $30^{\circ}$ west of north

$v_{br}=-8cos30\hat{i}+8sin30\hat{j}$

$v_{r}=2\hat{i}$

Therefore velocity of boat w.r.t ground

$v_b=v_{br}+v_r=-8cos30\hat{i}+8sin30\hat{j}+2\hat{i}$

$v_b=-4.928\hat{i}+4\hat{j}$

Magnitude of boat speed$=\sqrt{4.928^2+4^2}=6.34 m/s$

For direction

$tan\theta =\frac{4}{4.928}$

$\theta =39.06^{\circ}$ w.r.t to west

to cross the river its north component will help so

time taken $t=\frac{200}{4}=50 s$

Answer 2

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.

$\vec v_{river}$ = 2.0·i

$\vec {v}_{boat}$ = -8.0·sin(30°)·i + 8.0·cos(30°)·j = -4.0·i + 4.0·√3·j

Therefore;

$\vec v_{resultant}$ = 2.0·i - 4.0·i + 4.0·√3·j = -2.0·i + 4.0·√3·j

$| \vec {v}_{resultant}|$ = v =  $\sqrt{(-2)^2 + \left(4.0\cdot \sqrt{3} \right) } = \mathbf{ 2\cdot \sqrt{13}}$

• The magnitude of the (resultant) velocity of the boat, v = $2 \cdot \sqrt{13}$ m/s ≈ 7.2 m/s

(b) Let θ represent the direction of the boat's relative velocity to the ground, we have;

$tan\left(\theta \right) = \mathbf{\dfrac{4.0 \cdot \sqrt{3} }{-2.0}}$

Therefore;

$\theta = arctan \left(\dfrac{4.0 \cdot \sqrt{3} }{-2.0} \right) \approx \mathbf{ -73.9^{\circ}}$

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

$Time = \mathbf{ \dfrac{Distance}{Velocity}}$

Time it takes the boat to cross the river is therefore;

• $Time = \dfrac{200 \, m}{4.0 \cdot \sqrt{3} \, m/s} = \dfrac{50 \cdot \sqrt{3} }{3} \, s \approx \underline{28.9 \, s}$

 

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