Aussie Maths & Science Teachers: Save your time with SmarterEd
The acceleration, \(a\) ms\(^{-2}\), of a particle that starts from rest and moves in a straight line is described by \(a=1+v\), where \(v\) ms\(^{-1}\) is its velocity after \(t\) seconds.
Determine the velocity of the particle after \( \log _e(e+1) \) seconds. (3 marks)
\(e\ \text{ms}^{-1}\)
\(\dfrac{dv}{dt}=1+v\ \ \Rightarrow \ \dfrac{dt}{dv} = \dfrac{1}{1+v} \)
\(t= \displaystyle{\int \dfrac{1}{1+v}\ dv} = \log_e{(1+v)}+c \)
\(\text{When}\ \ t=0, v=0\ \ \Rightarrow \ c=0 \)
| \(t\) | \(=\log_e(1+v) \) | |
| \(1+v\) | \(=e^t\) | |
| \(v\) | \(=e^t-1\) |
\(\text{At}\ \ t=\log_e(e+1): \)
\(v=e^{\log_e{(e+1)}}-1 = e+1-1=e\ \text{ms}^{-1} \)
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)`
| `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)`
The acceleration, `a\ text(ms)^(-2)`, of a particle moving in a straight line is given by `a = v^2 + 1`, where `v` is the velocity of the particle at any time `t`. The initial velocity of the particle when at origin O is `2\ text(ms)^(-1)`.
The displacement of the particle from O when its velocity is `3\ text(ms)^(-1)` is
`C`
| `v* (dv)/(dx)` | `= v^2 + 1` |
| `(dv)/(dx)` | `= (v^2 + 1)/v` |
| `(dx)/(dv)` | `= v/(v^2 + 1)` |
| `x` | `= int (v/(v^2 + 1))\ dv` |
| `=1/2 ln(v^2 + 1)+c` |
`text(When)\ \ x=0, \ v=2:`
`c=-1/2 ln5`
`text(Find)\ \ x\ \ text(when)\ \ v=3:`
| `x` | `= 1/2ln(9 + 1) – 1/2 ln5` |
| `= 1/2 ln (10/5)` | |
| `= 1/2 ln 2` |
`=> C`
The acceleration `a` ms¯² of a body moving in a straight line in terms of the velocity `v` ms¯¹ is given by `a = 4v^2.`
Given that `v = e` when `x = 1`, where `x` is the displacement of the body in metres, find the velocity of the body when `x = 2.` (4 marks)
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`e^5`
| `v* (dv)/(dx)` | `= 4v^2` |
| `(dv)/(dx)` | `= 4v` |
| `(dx)/(dv)` | `= 1/(4v)` |
| `x` | `=int 1/(4v) \ dv` |
| `=1/4 ln (4v)+c` |
`text(When)\ \ x=1, v=e`
| `1` | `=1/4 ln(4e)+c` | |
| `:.c` | `=1-1/4 ln(4e)` |
`text(Find)\ \ v\ \ text(when)\ \ x=2:`
COMMENT: Strong log and exponential calculation ability required here.
| `1/4 ln (4v)+1-1/4 ln(4e)` | `= 2` |
| `1/4 ln(4v)` | `= 1/4 ln (4e)+1` |
| `ln(4v)` | `= ln(4e)+4` |
| `e^(ln4v)` | `= e^(ln(4e) +4)` |
| `4v` | `= e^(ln(4e)) * e^4` |
| `4v` | `=4e xx e^4` |
| `:.v` | `=e^5` |
A particle starts at the origin with velocity 1 and acceleration given by
`a = v^2 + v`,
where `v` is the velocity of the particle.
Find an expression for `x`, the displacement of the particle, in terms of `v`. (3 marks)
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`x= ln \ | frac{v + 1}{2} |`
| `a` | `= v^2 + v` |
| `v · frac{dv}{dx}` | `= v^2 + v` |
| `frac{dv}{dx}` | `= v + 1` |
| `frac{dx}{dv}` | `= frac{1}{v + 1}` |
| `x` | `= int frac{1}{v + 1}\ dv` |
| `= ln \ | v + 1 | + c` |
`text{When} \ \ x = 0 , \ v = 1`
`0 = ln \ 2 + c`
`c = -ln \ 2`
| `therefore \ x` | `= ln \ | v + 1| – ln \ 2` |
| `= ln \ | frac{v + 1}{2} |` |