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PHYSICS, M1 EQ-Bank 17

A boy on a bike is travelling at 8 ms\(^{-1}\) north when he sees a girl on a scooter travelling at 6 ms\(^{-1}\) west, relative to the ground. 

  1. Draw a vector diagram to represent the velocity of the girl relative to the boy.   (2 marks)

  2. Use the vector diagram to calculate the velocity of the girl relative to the boy.   (2 marks)

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10 ms\(^{-1}\), S36.9\(^{\circ}\)W

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a.    \(v_{\text{g}} = 6\ \text{ms}^{-1}\ \text{west} \)

\(-v_{\text{b}} = 8\ \text{ms}^{-1}\ \text{south} \)

\(v_{\text{g rel b}} = v_\text{g} + (-v_\text{b}) \)
 


 

b.    \(v_{\text{g rel b}}=\sqrt{6^2+8^2} =10\ \text{ms}^{-1}\)

  \(\tan \theta\) \(=\dfrac{8}{6}\)
  \(\theta\) \(=\tan^{-1}\Big(\dfrac{8}{6}\Big)=53.1^{\circ}\)

 
\(\therefore\) Velocity of the girl relative to the boy is 10 ms\(^{-1}\), S36.9°W.

PHYSICS, M1 EQ-Bank 16

Car A is travelling north at 40 ms\(^{-1}\) and Car B is travelling at 30 ms\(^{-1}\) east.

Determine the velocity of Car A relative to Car B.   (3 mark)

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50 ms\(^{-1}\), N36.9\(^{\circ}\)W

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\(v_{\text{A rel B}}=v_A-v_B =v_A + (-v_B)\) 

\(v_A = 40\ \text{ms}^{-1}\ \text{north}, \ -v_B = 30\ \text{ms}^{-1}\ \text{west} \)

\(v_{\text{A rel B}}=\sqrt{40^2+30^2}=50\ \text{ms}^{-1}\)

\(\tan \theta\) \(=\dfrac{30}{40}\)  
\(\theta\) \(=\tan^{-1}\Big(\dfrac{30}{40}\Big)=36.9^{\circ}\)  

 

\(\terefore\) Velocity of Car A relative to Car B is 50 ms\(^{-1}\), N36.9°W.

PHYSICS, M1 EQ-Bank 5

Car A approaches an intersection at 10 ms\(^{-1}\) from the South as shown. As Car B approaches the intersection, it measures the velocity of Car A to be 22.36 ms\(^{-1}\) NW.
 

Draw Car B on the diagram showing its direction and speed. Show all working.  (3 marks)

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→ Velocity of A relative to B = \(v_A-v_B=v_A+-v_B\)
 

→ The direction of \(-v_B\) must be west, and therefore the direction of \(v_B\) is east.

→ Using Pythagoras theorem, the magnitude of \(-v_B\) is:

\(-v_B\) \(=\sqrt{22.36^2-10^2}\)  
  \(=20\ \text{ms}^{-1}\)  

 

→ Thus the vector \(v_B\) is \(20\ \text{ms}^{-1}\) east.
 

Show Worked Solution

→ Velocity of A relative to B = \(v_A-v_B=v_A+-v_B\)
  

→ The direction of \(-v_B\) must be west, and therefore the direction of \(v_B\) is east.

→ Using Pythagoras theorem, the magnitude of \(-v_B\) is:

\(-v_B\) \(=\sqrt{22.36^2-10^2}\)  
  \(=20\ \text{ms}^{-1}\)  

 

→ Thus the vector \(v_B\) is \(20\ \text{ms}^{-1}\) east.
 

PHYSICS, M1 EQ-Bank 4

A plane that can fly at 500 kmh\(^{-1}\) with no wind, encounters a strong cross wind of 100 kmh\(^{-1}\) from the east. The plane needs to travel directly north to an airstrip

  1. Calculate the angle, correct to one decimal place, at which the pilot should steer for the plane to fly directly to the airstrip?   (3 marks)

  2. If the airstrip is 100 km from the planes current position, how long, to the nearest minute, will it take for the plane to complete the journey?   (2 marks)

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i.    \(\text{N11.5°E}\)

ii.    \(\text{12 minutes}\)

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i.    
       

\(\sin\theta\) \(=\dfrac{100}{500}\)  
\(\theta\) \(=\sin^{-1}\Big{(}\dfrac{100}{500}\Big{)}\)  
  \(=11.5^{\circ}\)  

 
\(\text{Plane direction should be N11.5°E}\)
 

ii.    \(\text{Using the diagram in part (i):}\)

\(\tan(11.5°)\) \(=\dfrac{100}{x}\)  
\(x\) \(=\dfrac{100}{\tan(11.5°)}\)  
  \(=491.5\ \text{kmh}^{-1}\)  

 

\(t\) \(=\dfrac{\text{distance}}{\text{speed}}\)  
  \(=\dfrac{100}{491.5}\)  
  \(=0.203\ \text{h}\)  
  \(=12\ \text{m (nearest minute)}\)