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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 2023 SPEC1 8

A body moves in a straight line so that when its displacement from a fixed origin `O` is `x` metres, its acceleration, `a`, is `-4 x \ text{ms}^{-2}`. The body accelerates from rest and its velocity, `v`, is equal to `-2 \ text{ms}^{-1}` as it passes through the origin. The body then comes to rest again.

Find `v` in terms of `x` for this interval.   (4 marks)

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` – 2sqrt(1-x^2)`

Show Worked Solution

`a = -4x`

`d/dx(1/2v^2)` `= -4x`  
`1/2v^2` `= -2x^2 +c`  

 
`v= -2\ \ \text{when}\ \ x = 0\ \ =>\ \ c = 2`

`v^2` `= -4x^2 + 4`  
`v^2` `= 4(1-x^2)`  

 
`v= -2\ \ text{when}\ \ x=0:`

`:.\ v` `= -sqrt(4(1-x^2))`  
  `= -2sqrt(1-x^2)`  

Mechanics, EXT2 M1 EQ-Bank 21

The acceleration, \(a\) ms\(^{-2}\), of a particle that starts from rest and moves in a straight line is described by  \(a=1+v\), where \(v\) ms\(^{-1}\) is its velocity after \(t\) seconds.

Determine the velocity of the particle after \( \log _e(e+1) \) seconds.   (3 marks)

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\(e\ \text{ms}^{-1}\)

Show Worked Solution

\(\dfrac{dv}{dt}=1+v\ \ \Rightarrow \ \dfrac{dt}{dv} = \dfrac{1}{1+v} \)

\(t= \displaystyle{\int \dfrac{1}{1+v}\ dv} = \log_e{(1+v)}+c \)

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

\(t\) \(=\log_e(1+v) \)  
\(1+v\) \(=e^t\)  
\(v\) \(=e^t-1\)  

 
\(\text{At}\ \ t=\log_e(e+1): \)

\(v=e^{\log_e{(e+1)}}-1 = e+1-1=e\ \text{ms}^{-1} \)

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 2013 SPEC2 18*

A particle moves in a straight line such that its acceleration is given by  `a = sqrt(v^2-1)` , where `v` is its velocity and `x` is its displacement from a fixed point.

Given that  `v = sqrt2`  when  `x = 0`, find the velocity `v` in terms of `x`.   (4 marks)

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`v = sqrt(1 + (1 + x)^2)`

Show Worked Solution
`v(dv)/(dx)` `= sqrt(v^2-1)`
`(dv)/(dx)` `= sqrt(v^2-1)/v`
`(dx)/(dv)` `= v/sqrt(v^2-1)`
`x` `=int v/sqrt(v^2-1)\ dv`

 
`text(Integration by substitution:)`

`text(Let)\ \ u=v^2-1`

`(du)/(dv) = 2v \ => \ 1/2 du = v\ dv`

`x` `= 1/2 int u^(- 1/2) \ du`
  `= 1/2 * 2 u^(1/2) + c`
  `= sqrt(v^2-1) + c`

 
`text(When)\ \ x=0, \ v = sqrt2\ \ =>\ \ c = −1`

`x` `= sqrt(v^2-1) -1`
`v^2` `= 1 + (1 + x)^2`

 
`:. v = sqrt(1 + (1 + x)^2)`