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Calculus, EXT1* C1 2017 HSC 15c

Two particles move along the `x`-axis.

When  `t = 0`, particle `P_1` is at the origin and moving with velocity 3.

For  `t >= 0`, particle `P_1` has acceleration given by  `a_1 = 6t + e^(-t)`.

  1. Show that the velocity of particle `P_1` is given by  `v_1 = 3t^2 + 4-e^(-t)`  (2 marks)

When  `t = 0`, particle `P_2` is also at the origin.

For  `t >= 0`, particle `P_2` has velocity given by  `v_2 = 6t + 1-e^(-t)`.

  1. When do the two particles have the same velocity?  (2 marks)

  2. Show that the two particles do not meet for  `t > 0`.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `t = 1`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.   `a_1` `= 6t + e^(-t)`
  `v_1` `= int a_1\ dt`
    `= int 6t + e^(-t)\ dt`
    `= 3t^2-e^(-t) + c`

 

`text(When)\ t = 0,\ v_1 = 3`

`3` `= 0-1 + c`
`c` `= 4`
`:. v_1` `= 3t^2 + 4-e^(-t) …\ text(as required)`

 

ii.  `v_2 = 6t + 1-e^(-t)`

`text(Find)\ \ t\ \ text(when)\ \ v_1 = v_2`

`3t^2 + 4-e^(-t)` `= 6t + 1-e^(-t)`
`3t^2-6t + 3` `= 0`
`t^2-2t + 1` `= 0`
`(t-1)^2` `= 0`
`:. t` `=1`

 

iii.   `x_1` `= int v_1\ dt`
    `= int 3t^2 + 4-e^(-t)\ dt`
    `= t^3 + 4t + e^(-t) + c`

 

`text(When)\ \ t = 0,\ \ x_1 = 0`

♦ Mean mark (iii) 39%.
`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_1` `= t^3 + 4t + e^(-t)-1`

 

`x_2` `= int 6t + 1-e^(-t)\ dt`
  `= 3t^2 + t + e^(-t) + c`

 
`text(When)\ \ t = 0,\ \ x_2 = 0`

`0` `= 0 + 0 + 1 + c`
`c` `= -1`
`:. x_2` `= 3t^2 + t + e^(-t)-1`

 

`text(Find)\ \ t\ \ text(when)\ \ x_1 = x_2`

`t^3 + 4t + e^(-t)-1` `= 3t^2 + t + e^(-t)-1`
`t^3-3t^2 + 3t` `= 0`
`t(t^2-3t + 3)` `= 0`

 

`text(S)text(ince)\ \ Delta < 0\ \ text(for)\ \ t^2-3t + 3`

`=>\ text(No real solution)`

 

`:.\ text(The particles do not meet)`

`(x_1 != x_2)\ \ text(for)\ \ t > 0.`