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Calculus, EXT1* C1 2007 HSC 5b

A particle is moving on the `x`-axis and is initially at the origin. Its velocity, `v` metres per second, at time `t` seconds is given by

`v = (2t)/(16 + t^2).`

  1. What is the initial velocity of the particle?  (1 mark)

  2. Find an expression for the acceleration of the particle.  (2 marks)

  3. Find the time when the acceleration of the particle is zero.  (1 mark)

  4. Find the position of the particle when `t = 4`.  (3 marks)

Show Answers Only
  1. `0`
  2. `{2(16 – t^2)}/(16 + t^2)^2`
  3. `4\ text(seconds)`
  4. `log_e 2\ \ text(metres)`
Show Worked Solution

i.  `v = (2t)/(16 + t^2)`

`text(When)\ t` `= 0`
`v` `= 0`

`:.\ text(Initial velocity is 0.)`

 

ii.  `a = d/(dt) ((2t)/(16 + t^2))`
 

`text(Using quotient rule)`

`u` `= 2t` `v` `= 16 + t^2`
`u prime` `= 2` `v prime` `= 2t`
`(dv)/(dt)` `= (u prime v – uv prime)/v^2`
  `= {2(16 + t^2) – 2t * 2t}/(16 + t^2)^2`
  `= (32 + 2t^2 – 4t^2)/(16 + t^2)^2`
  `= {2(16 – t^2)}/(16 + t^2)^2`

 

iii.  `text(Find)\ t\ text(when)\ (dv)/(dt) = 0`

`{2 (16 – t^2)}/(16 + t^2)^2` `= 0`
`2 (16 – t^2)` `= 0`
`t^2` `= 16`
`t` `= 4\ ,\ t >= 0`

 

`:.\ text(The acceleration is zero when)`

`t = 4\ text(seconds.)`

 

iv.  `v = (2t)/(16 + t^2)`

`x` `= int v\ dt`
  `= int (2t)/(16 + t^2)`
  `= log_e (16 + t^2) + c`

 

`text(When)\ \ t = 0\ ,\ x = 0`

`0 = log_e (16 + 0) + c`

`c = -log_e 16`

`:. x = log_e(16 + t^2) – log_e 16`
 

`text(When)\ t = 4,`

`x` `= log_e (16 + 4^2) – log_e 16`
  `= log_e 32 – log_e 16`
  `= log_e (32/16)`
  `= log_e 2\ \ text(metres)`

 

`:.\ text(When)\ t = 4\ , \ text(the position of the)`

`text(particle is)\ log_e 2 \ text(metres.)`