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Mechanics, EXT2 M1 2025 SPEC1 3

A particle starts from rest at a fixed point \(O\) and travels in a straight line.

The velocity, \(v\) m s\(^{-1}\), of the particle at time \(t\) seconds has equation  \(v(t)=\dfrac{t}{\sqrt{t^2+k}}\), where \(k\) is a positive constant and  \(t \geq 0\).

  1. Use integration to show that the displacement, \(x\) metres, of the particle relative to \(O\) is given by  \(x(t)=\sqrt{t^2+k}-\sqrt{k}\).   (1 mark)

  2. Find the initial acceleration, in terms of \(k\), of the particle in m s\(^{-2}\).   (2 marks)

  3. Another particle starts at \(O\) at the same time as the first particle and follows the same path.
  4. Its position relative to \(O\) is described by the equation  \(s(t)=t\).
  5. Three seconds after leaving \(O\) the second particle is 1 m ahead of the first particle.
  6. Find the value of \(k\).   (2 marks)

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a.    \(v=\dfrac{t}{\sqrt{t^2+k}}\)

\(x=\displaystyle \int \frac{t}{\sqrt{t^2+k}} d t=\sqrt{t^2+k}+c\)
 

\(\text{When} \ \  t=0, x=0:\)

\(0=\sqrt{k}+c \ \ \Rightarrow \ \ c=-\sqrt{k}\)

\(\therefore x=\sqrt{t^2+k}-\sqrt{k}\)
 

b.    \(a=-\sqrt{k}\)

c.    \(k=\dfrac{25}{16}\)

Show Worked Solution

a.    \(v=\dfrac{t}{\sqrt{t^2+k}}\)

\(x=\displaystyle \int \frac{t}{\sqrt{t^2+k}} d t=\sqrt{t^2+k}+c\)
 

\(\text{When} \ \  t=0, x=0:\)

\(0=\sqrt{k}+c \ \ \Rightarrow \ \ c=-\sqrt{k}\)

\(\therefore x=\sqrt{t^2+k}-\sqrt{k}\)
 

b.    \(\text{Find initial acceleration:}\)

\(v=t\left(t^2+k\right)^{-\tfrac{1}{2}}\)

\(\text{Using product rule:}\)

\(\dfrac{dv}{dt}\) \(=t\cdot-\dfrac{1}{2} \cdot 2 t\left(t^2+k\right)^{-\tfrac{3}{2}}+\left(t^2+k\right)^{-\tfrac{1}{2}}\)
  \(=-t^2\left(t^2+k\right)^{-\tfrac{3}{2}}+\left(t^2+k\right)^{-\tfrac{1}{2}}\)

 

\(\text{At} \ \ t=0:\)

\(a=\dfrac{dv}{dt}=0+k^{-\tfrac{1}{2}}=-\sqrt{k}\)
 

c.   \(\text{1st particle:} \ \ x(3)=\sqrt{9+k}-\sqrt{k}\)

\(\text{2nd particle:} \ \ s(3)=3\)
 

\(\text{Since at \(t=3\), second particle is 1m ahead:}\)

\(3-\sqrt{9+k}+\sqrt{k}=1\)

\(\sqrt{9+k}\) \(=2+\sqrt{k}\)
\(\left(\sqrt{9+k}\right)^2\) \(=(2+\sqrt{k})^2\)
\(9+k\) \(=4+4 \sqrt{k}+k\)
\(4 \sqrt{k}\) \(=5\)
\(\sqrt{k}\) \(=\dfrac{5}{4}\)
\(k\) \(=\dfrac{25}{16}\)
♦ Mean mark (c) 40%.

Mechanics, EXT2 M1 EQ-Bank 22

A student throws a ball for his dog to retrieve. The position vector of the ball, relative to an origin \(O\) at ground level \(t\) seconds after release, is given by  \(  \underset{\sim}{\text{r}}{}_\text{B} (t)=5 t \underset{\sim}{\text{i}}+7 t \underset{\sim}{\text{j}}+(15 t-4.9 t^2+1.5) \underset{\sim}{\text{k}} \). Displacement components are measured in metres, where \(\underset{\sim}{\text{i}}\) is a unit vector to the east, \(\underset{\sim}{\text{j}}\) is a unit vector to the north and \( \underset{\sim} {\text{k}}\) is a unit vector vertically up.

Calculate the total vertical distance, in metres, travelled by the ball before it hits the ground. Give your answer correct to one decimal place.   (3 marks)

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\(24.5\ \text{m} \)

Show Worked Solution

\(\text{Upwards distance}\ (z) = 15t-4.9t^2+1.5\)

\(\dfrac{dz}{dt}=15-9.8t\)

\(\text{Find}\ t\ \text{when}\ \dfrac{dz}{dt}=0\ \text{(vertical max):} \)

\(15-9.8t=0\ \ \Rightarrow \ \ t=\dfrac{15}{9.8} \)

\(z\Big{(}t=\frac{15}{9.8}\Big{)} = 15 \times \Big{(}\dfrac{15}{9.8}\Big{)}-4.9 \times \Big{(}\dfrac{15}{9.8}\Big{)}^2 + 1.5 \approx 12.98\ \text{m} \)
  

\(\text{At}\ \ t=0, \ z=1.5 \)

\(\therefore \text{Total vertical distance}\ = (12.98-1.5)+12.98 = 24.5\ \text{m (1 d.p.)} \)

Mechanics, EXT2 M1 2023 HSC 12c

An object with mass \(m\) kilograms slides down a smooth inclined plane with velocity \( \underset{\sim}{v}(t)\), where \(t\) is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle \(\theta\) with the horizontal, as shown in the diagram. The normal reaction force is \(\underset{\sim}{R}\). The acceleration due to gravity is \(\underset{\sim}{g}\) and has magnitude \(g\). No other forces act on the object.

The vectors \(\underset{\sim}{i}\) and \( \underset{\sim}{j} \) are unit vectors parallel and perpendicular, respectively, to the plane, as shown in the diagram.
 

  1. Show that the resultant force on the object is  \(\underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i}\).   (2 marks)

  2. Given that the object is initially at rest, find its velocity \(\underset{\sim}{v}(t)\) in terms of \(g\), \(\theta, t\) and \(\underset{\sim}{i}\).   (2 marks)

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i.    \(\text{Proof (See Worked Solution)} \)

ii.   \(\underset{\sim}{v}=-gt\ \sin \theta \ \underset{\sim}{i} \)

Show Worked Solution

i.       
         

\(\text{Resolving forces in}\ \underset{\sim}{j} \ \text{direction:} \)

\( {\underset{\sim}{F}}_\underset{\sim}{j} = \underset{\sim}{R} + m\underset{\sim}{g}\ \cos \theta = 0\ \ \text{(in equilibrium)} \)

\(\text{Resolving forces in}\ \underset{\sim}{i} \ \text{direction:} \)

\( {\underset{\sim}{F}}_\underset{\sim}{i} = -m\underset{\sim}{g} \ \sin \theta \ \ \ \text{(down slope)} \)

\(\therefore \text{Resultant force:}\ \ \underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i} \)
 

♦ Mean mark (i) 50%.

ii.   \(\text{Using}\ \ \underset{\sim}{F}=m \underset{\sim}{a}: \)

\(m \underset{\sim}{a}\) \(=-mg\ \sin \theta \ \underset{\sim}{i} \)  
\(\underset{\sim}{a}\) \(=-g\ \sin \theta \ \underset{\sim}{i} \)  
\(\underset{\sim}{v}\) \(= \displaystyle \int \underset{\sim}{a}\ dt \)  
  \(=-gt\ \sin \theta +c \)  

 
\(\text{When}\ \ t=0,\ \ \underset{\sim}{v}=0\ \ \Rightarrow \ \ c=0 \)

\(\therefore \underset{\sim}{v}=-gt\ \sin \theta \ \underset{\sim}{i} \)

Mechanics, EXT2 M1 2022 HSC 12b

A particle is moving in a straight line with acceleration  `a=12-6 t`. The particle starts from rest at the origin.

What is the position of the particle when it reaches its maximum velocity?   (3 marks)

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`x=16`

Show Worked Solution

`a=12-6t`

`v=int 12-6t\ dt=12t-3t^2+c`

  
`text{When}\ \ t=0, v=0\ \ =>\ \ c=0`

`x=int 12t-3t^2\ dt=6t^2-t^3+c`

 
`text{When}\ \ t=0, x=0\ \ =>\ \ c=0`

`v_max\ \ text{occurs when}\ \ a=0:`

`12-6t=0\ \ =>\ \ t=2`

`:.x|_(t=2)=6(2^2)-2^3=16`

Mechanics, EXT2* M1 2007 HSC 2d

A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, `v` metres per second, at which she is falling `t` seconds after jumping is given by  `v =50(1-e^(-0.2t))`.

  1. Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place.   (2 marks)

  2. Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre.   (2 marks)

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a.    `1.4\ text(ms)^(−2)\ \ text{(to 1 d.p.)}`

b.    `284\ text{m  (nearest m)}`

Show Worked Solution
a.      `v` `= 50(1 − e^(-0.2t))`
    `=50-50e^(-0.2t)`
  `ddot x` `= (dv)/(dt)`
    `= −0.2 xx 50 xx −e^(−0.2t)`
    `= 10 e^(−0.2t)`

 
`text(When)\ \ t = 10:`

`ddot x` `= 10 e^(−0.2 xx 10)`
  `= 10 e ^(−2)`
  `= 1.353…`
  `= 1.4\ text(ms)^(−2)\ \ \ text{(to 1 d.p.)}`

 

b.    `text(Distance travelled)`

`= int_0^10 v\ dt`

`= 50 int_0^10 1 − e^(−0.2t) \ dt`

`= 50 [t + 1/0.2 · e^(−0.2t)]_0^10`

`= 50 [t + 5e^(−0.2t)]_0^10`

`= 50 [(10 + 5e^(−2)) − (0 + 5e^0)]`

`= 50 [10 + 5e^(−2) − 5]`

`= 50 [5 + 5e^(−2)]`

`= 283.833…`

`= 284\ text{m  (nearest m)}`