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PHYSICS, M1 EQ-Bank 14

A boat is trying to cross a river that is 200 meters wide. The boat’s speed in still water is 5 m/s. The river flows eastward with a current of 3 m/s.

  1. If the boat heads straight north (perpendicular to the current), using a vector diagram. determine the magnitude and direction of its velocity relative to the ground?   (3 marks)

  1. How far downstream (east) will the boat have drifted by the time it reaches the opposite bank?   (1 mark)

  1. At what angle should the boat be aimed (relative to north) so that it reaches directly across from its starting point?   (2 marks)

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a.    The velocity of the boat relative to the ground is 5.83 m/s N31\(^{\circ}\)E.

b.    \(120\ \text{m}\)

c.    The boat should head N36.9\(^{\circ}\)W.

Show Worked Solution

a.    Using the vector diagrams below

\(x=\sqrt{3^2+5^2} = 5.83\ \text{ms}^{-1}\)

\(\tan \theta= \dfrac{3}{5}\ \ \Rightarrow\ \ \theta=\tan^{-1}\left(\dfrac{3}{5}\right)=31^{\circ}\)

 \(\therefore\) The velocity of the boat relative to the ground is 5.83 m/s N31\(^{\circ}\)E.
 

b.    
         

\(\tan 31^{\circ}\) \(=\dfrac{d}{200}\)  
\(d\) \(=200 \times \tan\,31^{\circ}=120\ \text{m}\)  

 
c.    
         

\(\sin \theta=\dfrac{3}{5}\ \ \Rightarrow\ \ \theta=\sin^{-1}\left(\dfrac{3}{5}\right)=36.9^{\circ}\)

 \(\therefore\) The boat should head N36.9\(^{\circ}\)W.

PHYSICS, M1 EQ-Bank 2 MC

A cyclist rides 500 m to the west and then turns to travel 1200 m to the north. What is the cyclist's final displacement?

  1. 1700 m
  2. 1700 m bearing N 23\(^{\circ}\) W
  3. 1300 m bearing N 67\(^{\circ}\) W
  4. 1300 m bearing N 23\(^{\circ}\) W
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\(D\)

Show Worked Solution

Using vector addition:

\(R=\sqrt{500^2+1200^2} = 1300\ \text{m}\)

\(\tan \theta\) \(=\dfrac{1200}{500}\)  
\(\theta\) \(=\tan^{-1}\left(\dfrac{1200}{500}\right)=67^{\circ}\)  

 
\(\Rightarrow D\)

PHYSICS, M1 EQ-Bank 13

A motorboat was travelling 15 m/s relative to the water, heading due east according to its navigation system.

The boat then experiences a river current flowing south at 5 m/s relative to the shore. Using a vector diagram, determine the boat's resultant velocity relative to the shore.   (2 marks)

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15.8 ms\(^{-1}\) at S72\(^{\circ}\)E.

Show Worked Solution

Using vector addition as shown in the diagram above:

\(R=\sqrt{15^2+5^2} = 15.8\ \text{ms}^{-1}\)

\(\tan \theta\) \(=\dfrac{5}{15}\)  
\(\theta\) \(=\tan^{-1}\left(\dfrac{5}{15}\right)=18.4^{\circ}\)  

 
The resultant velocity is 15.8 ms\(^{-1}\) at S72\(^{\circ}\)E.

PHYSICS, M1 EQ-Bank 12

An aircraft flies with a constant velocity of 95 ms\(^{-1}\) North. During the flight, a the plane experiences 2 different cross winds. 

  1. Determine the resultant velocity of the plane as seen from the ground when it experiences a wind blowing 40 ms\(^{-1}\) to the west.   (2 marks)
  1. Later in the flight when the plane is again travelling 95 ms\(^{-1}\) North, calculate its resultant velocity as seen from the ground when it experiences a cross wind of 50 ms\(^{-1}\) blowing towards a bearing of 150°T.   (4 marks)

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a.    \(103.1\ \text{ms}^{-1}\) at \(337^{\circ}\ \text{T}\)

b.    \(57.4\ \text{ms}^{-1}\) at \(026^{\circ}\ \text{T}\)

Show Worked Solution

a.   
       

Using vector addition:

\(R=\sqrt{40^2+95^2} = 103.1\ \text{ms}^{-1}\)

\(\tan \theta\) \(=\dfrac{40}{95}\)  
\(\theta\) \(=\tan^{-1}\left(\dfrac{40}{95}\right)=23^{\circ}\)  

 
Resultant velocity \(=103.1\ \text{ms}^{-1}\ \text{at}\ 337^{\circ}\ \text{T}\)
 

b. 
       

Using the cosine rule:

\(c=\sqrt{50^2+95^2-2 \times 50 \times 95 \times \cos 30} = 57.4261\ \text{ms}^{-1}\)
 

Determine the angle (using sine rule):

\(\dfrac{\sin \theta}{50}\) \(=\dfrac{\sin 30}{R}\)  
\(\sin \theta\) \(=\dfrac{50 \times \sin 30}{57.4261}\)  
\(\theta\) \(=\sin^{-1}\left(\dfrac{50 \times \sin 30}{57.4261}\right)=25.8^{\circ}\)   

 

Resultant velocity on the plane \(=57.4\ \text{ms}^{-1}\ \text{at}\ 26^{\circ}\ \text{T}\)

PHYSICS, M1 EQ-Bank 1 MC

A hiker starts at Camp A and walks 1.2 km east. She then turns to face north and walks a further 0.9 km to arrive at Camp B.

The magnitude of the hiker’s overall displacement \((d)\) can be determined using which of the following expressions?

  1. \(\tan d = \dfrac{0.9\ \text{km}}{1.2\ \text{km}}\)
  2. \(d = 1.2\ \text{km} + 0.9\ \text{km}\)
  3. \(d = (1.2)^2\ \text{km} + (0.9)^2\ \text{km}\)
  4. \(d^2 = (1.2)^2\ \text{km} + (0.9)^2\ \text{km}\)
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\(D\)

Show Worked Solution
  • By Pythagorean theorem:

\(d^2 = (1.2)^2\ \text{km} + (0.9)^2\ \text{km}\)

\(\Rightarrow D\)

PHYSICS, M1 EQ-Bank 1

A physics student comes across a river which runs north to south and has a current of 3 ms\(^{-1}\) running south.

The student starts on the west side of the river at point A and paddles a kayak at 5 ms\(^{-1}\) directly across the river to finish at point B.

  1. Calculate the angle which he must position the boat to travel in a straight line across the river.   (2 mark)

  2. If the river is 100 metres wide, determine the time it takes for the student to cross the river.   (2 mark)

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a.    \(36.9^{\circ}\)

b.    \(\text{25 seconds}\)

Show Worked Solution

a.   
       

\(\sin \theta\) \(=\dfrac{3}{5}\)  
\(\theta\) \(=\sin^{-1}\Big(\dfrac{3}{5}\Big)=36.9^{\circ}\)  

 
The student must turn 36.9\(^{\circ}\) into the current as shown on the diagram.

 
b.    Using Pythagoras:

\(v=\sqrt{5^2-3^2}=4\ \text{ms}^{-1}\)

\(t=\dfrac{d}{s}=\dfrac{100}{4}=25\ \text{s}\)
 

\(\therefore\) It will take the student 25 seconds to travel from A to B.

PHYSICS, M1 EQ-Bank 10

Determine the magnitude and direction of the resultant velocity vector for an airplane that is simultaneously moving with a velocity of 350\(^{-1}\) directly towards the west and 275 ms\(^{-1}\) towards the northwest.   (3 marks)

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578 ms\(^{-1}\), N70.3°W

Show Worked Solution

Add vectors to show resultant vector:

Using the cosine rule to find the magnitude of \(v\):

\(v\) \(=\sqrt{350^2+275^2-2 \times 350 \times 275 \times \cos 135°}\)  
  \(=578.1376\dots\ \text{ms}^{-1}\)  

 

Using the sign rule to find \(\theta\):

\(\dfrac{\sin \theta}{275}\) \(=\dfrac{\sin 135°}{578.1376}\)  
\(\sin \theta\) \(=\dfrac{\sin135° \times 275}{578.1376}\)  
\(\theta\) \(=\sin^{-1}\Big(\dfrac{\sin135° \times 275}{578.1376}\Big)\)  
  \(=19.7^{\circ}\)  

 

\(\therefore\) Resultant velocity of the aeroplane is 578 ms\(^{-1}\), N70.3°W.

PHYSICS, M1 EQ-Bank 9

A car is travelling north and approaching an intersection at 50 kmh\(^{-1}\).

While maintaining a constant speed, the car turns left and continues east at 50 kmh\(^{-1}\). 

Using a vector diagram, calculate the change in velocity of the car.   (3 marks)

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\(\text{70.7 kmh}^{-1}, \text{S45°E} \)

Show Worked Solution

\(v= 50\ \text{kmh}^{-1}\ \text{east},\ \ u= 50\ \text{kmh}^{-1}\ \text{north.}\)

\(\Delta v= v-u = v+(-u),\ \ \text{where}\ -u= 50\ \text{kmh}^{-1}\ \text{south}\)
  

\(\Delta v=\sqrt{50^2 + 50^2}=\sqrt{5000}=70.7\ \text{kmh}^{-1}\)

\(\tan \theta \) \(=\dfrac{50}{50}\)  
\(\theta\) \(=\tan^{-1}\Big(\dfrac{50}{50}\Big)=45^{\circ}\)  

 

Change in velocity of the car = 70.7 \(\text{kmh}^{-1}\), S45\(^{\circ}\)E.

PHYSICS, M1 EQ-Bank 6

A boat is being rowed due north at a constant speed of 15 ms\(^{-1}\) when it encounters a current of 8 ms\(^{-1}\) going in the direction of S25°W.

Using vectors, determine the resultant velocity of the boat.   (3 marks)

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\(\overset{\rightarrow}R=8.455\ \text{ms}^{-1}\ \text{N}23.6^{\circ}\text{W}\)

Show Worked Solution

  • Using the cosine rule, the magnitude of \(\overset{\rightarrow}R\) is:
  •    \(\overset{\rightarrow}R=\sqrt{15^2+8^2-2 \times 15 \times 8 \times \cos25°}=8.455\ \text{ms}^{-1}\)
  • Using the sine rule, the direction of \(\overset{\rightarrow}R\) is:
\(\dfrac{\sin\theta}{8}\) \(=\dfrac{\sin25°}{8.455}\)  
\(\sin\theta\) \(=\dfrac{8\sin25°}{8.455}\)  
\(\theta\) \(=\sin^{-1}\Big{(}\dfrac{8\sin25°}{8.455}\Big{)}=23.6^{\circ}\)  

 
\(\therefore \overset{\rightarrow}R =8.455\ \text{ms}^{-1}\ \text{N}23.6^{\circ}\text{W}\)

PHYSICS, M1 2018 VCE 5*

Four students are pulling on ropes in a four-person tug of war. The relative sizes of the forces acting on the various ropes are \(F_{ W }=200 \text{ N} , F_{ X }=240 \text{ N} , F_{ Y }=180 \text{ N}\) and \(F_{ Z }=210 \text{ N}\). The situation is shown in the diagram below.
 

What is the resultant force vector \((F_{\vec{R}})\) acting at the centre of the tug-of-war ropes?   (3 marks)

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\(F_{\vec{R}}=36.1\ \text{N}, 33.7^{\circ}\) above the horizontal.

Show Worked Solution
  • By resolving the vertical vectors of \(F_{ W }=200 \text{ N}\) up and \(F_{ Y }=180 \text{ N}\) down, the net force in the vertical direction is \(F_v=20\ \text{N}\) up.
  • Similarly, the net force in the horizontal direction is \(F_h=30\ \text{N}\) to the right.
  • Thus the resultant vector \((F_{\vec{R}})\) can be calculated using the vector diagram below.

  • The magnitude of  \(F_{\vec{R}}=\sqrt{20^2+30^2}\)  and  \(\theta=\tan^{-1}\Big{(}\dfrac{20}{30}\Big{)}\)
  • Hence \(F_{\vec{R}}=36.1\ \text{N}, 33.7^{\circ}\) above the horizontal.