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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 as shown in the diagram above:

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

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

 

\(\Rightarrow D\)

PHYSICS, M1 EQ-Bank 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 \(40\ \text{ms}^{-1}\) to the west.   (2 marks)
  1. Calculate the resultant velocity of the plane as seen from the ground when it experiences a cross wind of \(50\ \text{ms}^{-1}\) blowing towards a bearing of \(150^{\circ}\)T.   (4 marks)

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

b.    The resultant velocity of the plane is \(57.4\ \text{ms}^{-1}\) at \(026^{\circ}\)T.

Show Worked Solution

a.   
       

→ Using vector addition as shown in the diagram above:

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

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

 

→ The resultant velocity is \(103.1\ \text{ms}^{-1}\) at \(337^{\circ}\)T.

 

b. 
       

→ Using the cosine rule, \(c^2=a^2+b^2-2ab\cos C\) where \(C\) is the angle opposite the unknown side.

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

To determine the angle, use the sine rule:

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

 

→ The resultant velocity of the plane is \(57.4\ \text{ms}^{-1}\) at \(026^{\circ}\)T.