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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)`