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Calculus, 2ADV C1 2014 HSC 13c v1

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

 `x =2t -3/sqrt(t+1)`,

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. `2`
Show Worked Solution
a.    `x` `=2t -3/sqrt(t+1)`
    `=2t -3(t+1)^(-1/2)`

 

`dot x` `= 2\ -3(-1/2) (t+1)^(-3/2)`
  `= 2 + 3/(2(t+1)^(3/2))`

 

`ddot x` `= -(9/4)(t+1)^(-5/2)`
  `= – 9/(4sqrt((t+1)^5))`

 
`text(S)text(ince)\ \ t >= 0,`

`=> 1/sqrt((t+1)^5) > 0`

`=> – 9/(4sqrt((t+1)^5)) < 0`
 

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

 

b.    `text(Velocity)\ (dot x) = 2 + 3/(2(t+1)^(3/2))`

 
`text(As)\ t -> oo,\ 3/(2(t+1)^(3/2)) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 2`

Calculus, 2ADV C1 EO-Bank 11

A particle is moving along the `x`-axis. Its velocity `v` at time `t` is given by

`v = (t^2+4)/sqrt(3t+1)`  metres per second

Find the acceleration of the particle when  `t = 2`.

Express your answer as an exact value in its simplest form.  (3 marks)

Show Answers Only

` (16sqrt(7))/49\ \ text(ms)^(−2)`

Show Worked Solution

`v = (t^2+4)/sqrt(3t+1)`

`alpha` `= (dv)/(dt)`

`text(Using quotient rule:)`

`u=t^2+4,`     `v=(3t+1)^(1/2)`  
`u^{′} = 2t,`     `v^{′} = 3/2 (3t+1)^(-1/2)`  
     
`alpha` `= (u^{′} v-v^{′} u)/v^2`
  `= (2t (3t+1)^(1/2)-3/2(t^2+4) (3t+1)^(-1/2))/(3t+1)`

 
`text(When)\ \ t = 2,`

`alpha` `= (4(7)^(1/2)-12(7)^(-1/2))/(7)`
  `= (4sqrt(7))/7 -12/(7sqrt(7))`
  `= (28sqrt(7))/49-(12sqrt(7))/49`
  `= (16sqrt(7))/49\ \ text(ms)^(−2)`

Calculus, 2ADV C1 EO-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = t^3-4t^2 +5t + 6`     when   `t >= 0`

  1.  Calculate the velocity when  `t = 3`.  (1 mark)

  2.  When is the particle stationary?  (2 marks)

Show Answers Only

i.    `8\ text(ms)^(−1)`

ii.   `1, 5/3\ text(s)`

Show Worked Solution

i.   `x =t^3-4t^2 +5t + 6` 

`v = (dx)/(dt) = 3t^2-8t +5`

 
`text(When)\ t = 3,`

`v` `= 3 xx 3^2-8 · 3 +5`
  `= 8\ text(ms)^(−1)`

 

ii.   `text(Particle is stationary when)\ \ v = 0`

`3t^2-8t +5` `= 0`
`3t^2-3t-5t+5` `= 0`
`3t(t-1)-5(t-1)` `= 0`
`(t-1)(3t-5)` `= 0`
`t` `= 1, 5/3\ text(s)`

Calculus, 2ADV C1 2018 HSC 12d v1

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

`x = 1/4t^4 -t^3 -1/2t^2 +3t,`

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

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

  2. At which times is the particle stationary?  (2 marks)

  3. Find the position of the particle when the acceleration is `4/3`.  (2 marks)

Show Answers Only
  1. `3\ text(ms)^(-1)`
  2. `t = 1 or 3\ text(seconds)`
  3. `-329/324\ text(m)`
Show Worked Solution

i.    `x = 1/4t^4 -t^3 -1/2t^2 +3t`

`v = (dx)/(dt) = t^3-3t^2-t+3`
 

`text(Find)\ \ v\ \ text(when)\ \ t = 0:`

`v` `= 0 -3(0) -0+ 3`
  `= 3\ text(ms)^(-1)`

 

ii.  `text(Particle is stationary when)\ \ v = 0`

`t^3-3t^2-t+3` `=0`  
`t^2(t-3)-1(t-3)` `=0`  
`(t-3)(t^2-1)` `=0`  
`(t-3)(t-1)(t+1)` `=0`  

 

`t = 1 or 3\ text(seconds), t >= 0`
 

iii.  `a = (dv)/(dt) = 3t^2-6t-1`
 

`text(Find)\ \ t\ \ text(when)\ \ a = 4/3`

`3t^2-6t-1` `= 4/3`
`3t^2-6t-7/3` `= 0`
`9t^2-18t-7` `=0`
`(3t-7)(3t+1)` `=0`
`t` `=7/3`, `t >= 0`
`x(7/3)` `= 1/4(7/3)^4 -(7/3)^3 -1/2(7/3)^2 +3(7/3)`
  `= -329/324\ text(m)`