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Calculus, 2ADV C1 2006 HSC 8a

A particle is moving in a straight line. Its displacement, `x` metres, from the origin, `O`, at time `t` seconds, where  `t ≥ 0`, is given by  `x = 1 - 7/(t + 4)`.

  1. Find the initial displacement of the particle.  (1 mark)

  2. Find the velocity of the particle as it passes through the origin.  (3 marks)

  3. Show that the acceleration of the particle is always negative.  (1 mark)

  4. Sketch the graph of the displacement of the particle as a function of time.  (2 marks)

Show Answers Only
  1. `text(–3/4 m)`
  2. `1/7\ text(ms)^-1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4.  
Show Worked Solution

i.   `x = 1 – 7/(t + 4)`
 

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

`x` `= 1 – 7/4`
  `= -3/4 \ \ text(m)`

 
`:.\ text(Initial displacement is)\ 3/4\ text(metres to)`

`text(the left of the origin.)`

 

ii.  `x = 1 – 7/(t+4) = 1 – 7(t + 4)^-1`

`dot x` `= (-1)  -7(t + 4)^-2 xx d/(dt)(t + 4)`
  `= 7 (t + 4)^-2 xx 1`
  `= 7/(t + 4)^2`

 

`text(Find)\ t\ text(when)\ x = 0`

`0` `= 1 – 7/(t + 4)`
`7/(t + 4)` `= 1`
`7` `= (t + 4)`
`t` `= 3`

 

`text(When)\ t = 3`

`dot x` `= 7/(3 + 4)^2`
  `= 1/7\ text(ms)^-1`

 

`:.\ text(The velocity of the particle as it passes)`

`text(through the origin is)\ 1/7\ text(ms)^-1.`

 

iii.  `dot x` `= 7(t + 4)^-2`
`ddot x` `= (d dot x)/(dt) = -14 (t +4)^-3`

 
`text(Given)\ t >= 0`

`=>  (t + 4)^-3 >= 0`

`=> -14 (t + 4)^-3 <= 0`

`:. ddot x\ text(is always negative.)`
 

(iv)  2UA HSC 2006 8a

Calculus, 2ADV C1 2014 HSC 13c

The displacement of a particle moving along the  `x`-axis is given by

 `x = t - 1/(1 + t)`,

where  `x`  is the displacement from the origin in metres,  `t`  is the time in seconds, and  `t >= 0`.

  1. Show that the acceleration of the particle is always negative.    (2 marks)

  2. What value does the velocity approach as  `t`  increases indefinitely?    (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1`
Show Worked Solution
i.    `x` `= t\ – 1/(1 + t)`
    `= t\ – (1 + t)^(-1)`

 

`dot x` `= 1\ – (-1) (1 + t)^(-2)`
  `= 1 + 1/((1 + t)^2)`

 

`ddot x` `= -2(1 + t)^(-3)`
  `= – 2/((1 + t)^3)`

 
`text(S)text(ince)\ \ t >= 0,`

`=> -2/((1 + t)^3) < 0`
 

`:.\ text(Acceleration is always negative.)`

 

ii.    `text(Velocity)\ (dot x) = 1 + 1/((1 + t)^2)`

 
`text(As)\ t -> oo,\ 1/((1 + t)^2) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 1`