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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)

Show Answers Only

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 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)

Show Answers Only

`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)`