# Math Help - Finding Velocity with Respect to Time (IS FRUSTRATINLGY HARD)

1. ## Finding Velocity with Respect to Time (IS FRUSTRATINLGY HARD)

hey this is a question i have for Year 12 Maths C in Australia. It is irritating me no end. <line of text deleted by CaptainBlack>
here we go...

A particle of mass one unit moves in a straight line in the horizontal plane against a resistance v+av^2 , where a is a positive constant. If v is the velocity of the particle at time t , then dv/dt=-(v+av^2) represents the particle’s motion.
Given this particle is released from the origin, O, with an initial velocity of u, find the velocity of the particle v as a function of the time t .

As the value of t increases, what happens to the velocity of this particle? Would this value change if a is negative?

THANKS

2. Originally Posted by Ned Hunter
hey this is a question i have for Year 12 Maths C in Australia. It is irritating me no end. <line of text deleted by CaptainBlack>
here we go...

A particle of mass one unit moves in a straight line in the horizontal plane against a resistance v+av^2 , where a is a positive constant. If v is the velocity of the particle at time t , then dv/dt=-(v+av^2) represents the particle’s motion.
Given this particle is released from the origin, O, with an initial velocity of u, find the velocity of the particle v as a function of the time t .

As the value of t increases, what happens to the velocity of this particle? Would this value change if a is negative?

THANKS
The ODE

$
\frac{dv}{dt}=-(v+av^2)
$

is of variables seperable type so has solution:

$
\int_{v=u}^{v(t)} \frac{1}{v+a v^2} dv = -\int_0^t dt
$

The first integral is done by resolving the integrand into partial fractions,
and the second is $-t$.

That will give you $t$ as a function of $v$ which
you will need to rearrange to give $v$ as a function of $t$.

RonL

3. Originally Posted by Ned Hunter

As the value of t increases, what happens to the velocity of this particle? Would this value change if a is negative?
If the initial velocity $u$ is positive the acceleration is negative and so the particle slows down and as $t \to \infty,\ \ v \to 0$.

If the initial velocity is negative, the velocity will change until the acceleration is zero, that is $v+av^2=0$, that is $v=0$ or $v=-1/a$. The first of these will not occur as all trajectories starting with negative velocity are repelled by this solution. The second is an attractor, that is all trajectories of $v$ which start negative more towards it.

RonL

4. Originally Posted by CaptainBlack
The ODE

$
\frac{dv}{dt}=-(v+av^2)
$

is of variables seperable type so has solution:

$
\int_{v=u}^{v(t)} \frac{1}{v+a v^2} dv = -\int_0^t dt
$

The first integral is done by resolving the integrand into partial fractions,
and the second is $-t$.

That will give you $t$ as a function of $v$ which
you will need to rearrange to give $v$ as a function of $t$.

RonL
hey erm i dont quite get what you are saying here? do u think you could type the steps you took to achieve this? that would be a huge help. thanks.

Ned

5. Originally Posted by Ned Hunter
hey erm i dont quite get what you are saying here? do u think you could type the steps you took to achieve this? that would be a huge help. thanks.

Ned
Hello,

as the Captain pointed out you have to get the variable t at one side of the equation and the variable v at the other side:

$\frac{dv}{dt}=-(v+av^2)$ . Multiply by $dt$

$dv=-(v+av^2) \cdot dt$. Now divide by $(v+av^2)$

$\frac{1}{(v+av^2)} \cdot dv=-dt$. Now integrate both sides. You'll get the equation which CaptainBlack gave you.

6. ## thanks

hey thanks heaps for that. erm hate to be an irritating twerp but can you type it out? i dont get how to rearrange it to get v(t)

cheers,
ned

7. Originally Posted by Ned Hunter
hey thanks heaps for that. ... i dont get how to rearrange it to get v(t)
...
Hello,

CaptainBlack wrote:"The first integral is done by resolving the integrand into partial fractions, and the second is -t."
I'm going to show you what he meant:

$\int_{v=u}^{v(t)} \frac{1}{v+a v^2} dv = -\int_0^t dt$. I assume that the integral of the right side isn't a problem to you.

Take $\frac{1}{v+a v^2} = \frac{N}{v} + \frac{M}{1+av} = \frac{N+Nav+Mv}{v(1+av)}$. Thus you know that
$N = 1 \text{ and } Nav+Mv = 0$ (There isn't a summand in the numerator of the fraction which contains the variable v). Solve for N and M:
N = 1 and M = -a. Therefore your integral becomes:

$\int_{v=u}^{v(t)} \frac{1}{v} dv - \int_{v=u}^{v(t)} \frac{a}{1+a v} dv= -\int_0^t dt$

1. $\int \frac{1}{v} dv = \ln(v)$

2. $\int \frac{a}{1+a v} dv = \ln(1+av)$ Use integration by substitution here: u = 1+av; u' = a

3. $-\int dt = -t$

Without considering the bounds you have now:

$\ln(v) - \ln(1+av) = -t$

$\ln \left(\frac{v}{1+av} \right) = -t$. Now de-logarithmize(?) using the base $e$:

$\frac{v}{1+av} = e^{-t}$. Multiply both sides by the denominator:

$v = e^{-t} \cdot (1+av)~\Longrightarrow~ v = e^{-t} + a \cdot e^{-t} \cdot v$

$v(1-a\cdot e^{-t})=e^{-t}$

$v=\frac{e^{-t}}{1+a\cdot e^{-t}} =\frac{1}{e^t (1+a\cdot e^{-t})}= \frac{1}{e^t-a}$

I'll leave the rest for you