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Vectors, EXT1 V1 2025 HSC 13c

At time \(t\), a particle has position vector  \(\underset{\sim}{r}(t)=t \underset{\sim}{i}+\dfrac{t^2}{9} \underset{\sim}{j}\), velocity vector \(\underset{\sim}{v}(t)\) and acceleration vector \(\underset{\sim}{a}(t)\).

Find the time when the angle between \(\underset{\sim}{v}(t)\) and \(\underset{\sim}{a}(t)\) is \(\dfrac{\pi}{4}\).   (4 marks)

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\(t=\dfrac{9}{2}\)

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\(\underset{\sim}{r}(t)=t \underset{\sim}{i}+\dfrac{t^2}{9} \underset{\sim}{j}, \ \ \underset{\sim}{v}(t)=\underset{\sim}{i}+\dfrac{2}{9} t \underset{\sim}{j}, \ \ \underset{\sim}{a}(t)=\dfrac{2}{9} \underset{\sim}{j}\)

\(\underset{\sim}{v}=\displaystyle \binom{1}{\frac{2}{9} t}, \quad \underset{\sim}{a}=\displaystyle \binom{0}{\frac{2}{9}}\)

\(\abs{\underset{\sim}{v}}=\sqrt{1^2+\left(\dfrac{2}{9} t\right)^2}=\sqrt{1+\dfrac{4}{81} t^2}\)

\(\abs{\underset{\sim}{a}}=\sqrt{\left(\dfrac{2}{9}\right)^2}=\dfrac{2}{9}\)
 

\(\text{Angle between vectors}=\dfrac{\pi}{4}:\)

\(\cos \dfrac{\pi}{4}\) \(=\dfrac{1 \times 0+\dfrac{2}{9} t \times \dfrac{2}{9}}{\dfrac{2}{9} \times \sqrt{1+\dfrac{4}{81} t^2}}\)
\(\dfrac{1}{\sqrt{2}}\) \(=\dfrac{\dfrac{2}{9} t}{\sqrt{1+\dfrac{4}{81} t^2}}\)
\(\sqrt{1+\dfrac{4}{81} t^2}\) \(=\dfrac{2 \sqrt{2}}{9} t\)
\(1+\dfrac{4}{81} t^2\) \(=\dfrac{8}{81} t^2\)
\(81+4 t^2\) \(=8 t^2\)
\(4 t^2\) \(=81\)
\(t^2\) \(=\dfrac{81}{4}\)
\(t\) \(=\dfrac{9}{2} \quad(t>0)\)

Vectors, EXT1 V1 SM-Bank 31

A drone is set to fly west at 38 km/h.

A cross wind diverts its path so that it travels with a speed of 45 km/h in the direction shown below. 
 

Calculate the speed, to one decimal place, and the bearing of the cross wind, to the nearest degree.  (3 marks)

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`text{41.9 km/h on a bearing of 022°.}`

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`underset~u + underset~v` `= (-45 sin 30 \ , \ 45 cos 30)`
  `= (-22.5 \ , \ {45 sqrt3}/{2})`

 
`underset~u = (-38 , 0) \ , \ underset~v = (x , y)`

`((-38),(0)) + ((x), (y))` `= ((-22.5),({45 sqrt3}/{2}))`
`((x), (y))` `= ((15.5),({45 sqrt3}/{2}))`

 

`| underset~v|` `= sqrt{15.5^2 + {45^2 xx 3}/{4}}`  
  `= 41.94…`  
  `= 41.9 \ text{km/h}`  

 

`tan theta` `= {{45 sqrt3}/{2}}/{15.5} = 2.514`
`theta` `= 68.3^@`

 

`:.\ text(The crosswind has a speed of 41.9 km/h on a bearing of 022°.)`

Vectors, EXT1 V1 SM-Bank 30

A man attempts to swim north across a river at a speed of 4.5 km/h.

If the river's current is moving at 11 km/h due east, find the bearing  (nearest degree) and speed (to 1 decimal place) that the man will actually be swimming.  (3 marks)

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`text{11.9 km/h on a bearing of} \ 068^@`

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`underset~u = (0, 4.5)`

`underset~v = (11, 0)`

`underset~u  + underset~v = ((0),(4.5)) + ((11),(0)) = ((11),(4.5))`
  

`tan theta`  `= 11/4.5`
`theta` `= tan^{-1} (11/4.5)`
`theta` `= 67.75`
`theta` `~~ 068^@`
   
`| underset~u + underset~v |` `= sqrt{11^2 + 4.5^2}`
  `= 11.88 ..`
  `= 11.9 \ text{km/h}\ \ text{(1 d.p.)}`

 
`:. \ text{Swimmer’s speed is 11.9 km/h on a bearing of} \ 068^@`

Vectors, EXT1 V1 2021 HSC 14a

A plane needs to travel to a destination that is on a bearing of 063°. The engine is set to fly at a constant 175 km/h. However, there is a wind from the south with a constant speed of 42 km/h.

On what constant bearing, to the nearest degree, should the direction of the plane be set in order to reach the destination?   (3 marks)

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\(75^{\circ}\)

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♦♦ Mean mark 31%.

\(\text{Bearing = 063}^{\circ}\ \ \Rightarrow\ \ \angle ABC = 63^{\circ}\ \text{(alternate)}\)

\(\text{Using the sine rule:}\)

\(\dfrac{\sin \angle BAC}{42}\) \(=\dfrac{\sin\,63^{\circ}}{175}\)  
\(\sin \angle BAC\) \(=\dfrac{42 \times \sin\,63^{\circ}}{175}\)  
  \(=0.2138…\)  
\(\angle BAC\) \(=12.34…\)  
  \(=12^{\circ}\ \text{(nearest degree)}\)  

 
\(\therefore \text{Bearing to set plane on}\ = 63+12=75^{\circ}\)