SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Vectors, EXT1 EQ-Bank 34

The position vector of a particle at time \(t\) is given by  \(\mathbf{r}(t)=n e^{-2 t}\,\mathbf{i}-t^2\,\mathbf{j}\), where \(n\) is a positive constant.

Determine \(n\) if the particle's acceleration is perpendicular to its velocity when  \(t=\dfrac{1}{2}\).   (3 marks)

Show Answers Only

\(n=\dfrac{e}{2}\)

Show Worked Solution
\(\underset{\sim}{r}(t)\) \(=n e^{-2 t} \underset{\sim}{i}-t^2 \underset{\sim}{j}\)
\(\underset{\sim}{v}(t)\) \(=-2n e^{-2t} \underset{\sim}{i}-2 t\underset{\sim}{i} \ \ \Rightarrow \ \ v\left(\frac{1}{2}\right)=-2 ne^{-1} \underset{\sim}{i}-\underset{\sim}{j}\)
\(\underset{\sim}{a}(t)\) \(=4 ne^{-2t} \underset{\sim}{i}-2 \underset{\sim}{j} \ \ \Rightarrow \ \ a\left(\frac{1}{2}\right)=4ne^{-1} \underset{\sim}{i}-2 \underset{\sim}{j}\)
 

\(\text{Velocity}\perp \text{acceleration at}\ \  t=\dfrac{1}{2}:\)

   \(\displaystyle \binom{-\tfrac{2 n}{e}}{-1}\binom{\tfrac{4 n}{e}}{-2}=0\)

\(-\dfrac{8 n^2}{e^2}+2=0 \ \ \Rightarrow \ \ n^2=\dfrac{e^2}{4} \ \ \Rightarrow \ \ n=\dfrac{e}{2}\ \ (n\gt 0)\)

Vectors, EXT1 EQ-Bank 30

The position vector of a particle that is moving along a curve at time `t` is given by 

`\mathbf{r}(t) = 3 cos (t) \mathbf{i} + 4 sin (t) \mathbf{j}, \ t >= 0`.

Determine the first time when the speed of the particle is a minimum.   (3 marks)

Show Answers Only

`t_1 = pi/2`

Show Worked Solution

`underset ~v(t)= -3 sin(t) underset ~i + 4 cos (t) underset ~j`

`text{Since speed =}\ |underset ~v(t)|:`

`|underset ~v(t)|` `= sqrt (9 sin^2(t) + 16 cos^2(t))`
  `= sqrt (9 sin^2(t) + 9 cos^2(t) + 7 cos^2(t))`
  `= sqrt (9 + 7 cos^2(t))`

 
`text(Minimised speed occurs when)\ cos(t) = 0:`

`:.  t_1 = pi/2`

COMMENT:
Undertilde notation in answers is allowed even when boldface vector notation appears in the question.

Vectors, EXT1 EQ-Bank 4 MC

The position of a body is given by  `\mathbf{r} = 3\mathbf{i} + \mathbf{j} `  metres at a particular time. The body moves with constant velocity and two seconds later its displacement is  `−\mathbf{i} + 5\mathbf{j} `  metres.

The velocity, in m s−1, of the body is

  1. `2\mathbf{i} + 6\mathbf{j}`
  2. `−2\mathbf{i} + 2\mathbf{j}`
  3. `−4\mathbf{i} + 4\mathbf{j}`
  4. `4\mathbf{i}-4\mathbf{j}`
Show Answers Only

`B`

Show Worked Solution

`Delta \mathbf{r}= (−1-3)\mathbf{i} + (5-1)\mathbf{j}= −4\mathbf{i} + 4\mathbf{j}`

`\mathbf{v} = (Delta \mathbf{r})/(Delta \mathbf{t})= (−4\mathbf{i} + 4\mathbf{j})/(2-0)=-2\mathbf{i}+\mathbf{j}`

`=> B`

Vectors, EXT1 EQ-Bank 6 MC

The position of a particle at time `t` is given by  `\mathbf{r}(t) = (sqrt(t-2))\mathbf{i} + (2t)\mathbf{j}`  for  `t >= 2`.

The cartesian equation of the path of the particle is

  1. `y = 2x^2 + 4, \ \ \ \ \ x >= 2`
  2. `y = 2x^2 + 2, \ \ \ \ \ x >= 2`
  3. `y = sqrt((x-4)/2),\ \ \ x >= 2`
  4. `y = 2x^2 + 2,\ \ \ \ \ x >= 0`
Show Answers Only

`A`

Show Worked Solution

`underset ~r(t) = (sqrt(t-2)\underset ~i + (2t) underset ~j)`

`x = sqrt(t-2)`

`text(Given)\ \ t >= 2\ \ =>\ \ x >=0`

`y = 2t\ \ =>\ \ t = y/2`

`:. x` `= sqrt(y/2-2)`
`x^2` `= y/2-2`
`y/2` `= x^2 + 2`
`y` `= 2x^2 + 4`

 
`=> A`

Vectors, EXT1 EQ-Bank 5 MC

The acceleration vector of a particle that starts from rest is given by

`underset ~a(t) = −4 sin(2t) underset ~i + 20 cos (2t) underset ~j`, where `t >= 0`.

The velocity vector of the particle, `underset ~v(t)`, is given by

  1. `−8 cos(2t) underset ~i-40 sin(2t) underset ~j`
  2. `2 cos(2t) underset ~i + 10 sin(2t) underset ~j`
  3. `(8-8 cos(2t)) underset ~i-40 sin(2t) underset ~j`
  4. `(2 cos(2t)-2) underset ~i + 10 sin(2t) underset ~j`
Show Answers Only

`D`

Show Worked Solution

`underset ~v(t)= int underset ~a (t)\ dt= (2 cos (2t) + c_0) underset ~i + (10 sin (2t) + c_1) underset ~j`

`text(S)text(ince)\ \ v=0\ \ text(when)\ \ t=0:`

`0= (2 cos (0) + c_0) underset ~i + (10 sin (0) + c_1) underset ~j`

`0=(2 + c_0) underset ~i + c_1 underset ~j`

`=> c_0 = -2, \ \  c_1 = 0`
 

`:. underset ~v(t) = (2 cos (2t)-2) underset ~i + 10 sin(2t) underset ~j`

`=> D`

Vectors, EXT1 EQ-Bank 23

An object is travelling around a circular path of radius 5 m with position vector

\(\textbf{r} (t)=5 \cos \left(t^2\right)\textbf{i} +5 \sin \left(t^2\right)\textbf{j} \quad t \geq 0\)

where \(t\) is the time in seconds.

Find an expression for the speed of the object in terms of \(t\).   (2 marks)

Show Answers Only

\(\abs{\textbf{v} (t)}=10 t \ \text{ms}^{-1}\)

Show Worked Solution

\(\text {The velocity is given by}\)

\(\textbf{v}(t)=\dfrac{d}{d t} \textbf{r}(t)=-10 t\, \sin \left(t^2\right) \textbf{i} +10 t\, \cos \left(t^2\right) \textbf{j}\)

\(\text {The speed is the magnitude of the velocity:}\)

\(\abs{\textbf{v} (t)}\) \(=\sqrt{100 t^2\, \sin ^2\left(t^2\right)+100 t^2\, \cos ^2\left(t^2\right)}\)
  \(=10 t \sqrt{\sin ^2\left(t^2\right)+\cos ^2\left(t^2\right)}\)
  \(=10 t \ \text{m s}^{-1}\)

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)

Show Answers Only

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

Show Worked Solution

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

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

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

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

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

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

Vectors, EXT1 V1 EQ-Bank 25

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)

Show Answers Only

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

Show Worked Solution
`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 EQ-Bank 20

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)

Show Answers Only

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

Show Worked Solution

`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)= 67.75~~ 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)

Show Answers Only

\(75^{\circ}\)

Show Worked Solution

♦♦ Mean mark 31%.

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

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

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

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