# Finding Velocity from Acceleration

• alex_b93
In summary: Here's a possible approach:du/dt = g - (k/m)u 'minus' a force termTry withdu/dt = g - (k/m)u + F/mYou are essentially trying to add a constant force that would make the acceleration of the particle equal to g. If you can show that the two positions are the same then you have an expression for F.In summary, the conversation discusses a problem involving a particle falling through a viscous liquid under the influence of gravity. The particle experiences a drag force that is proportional to its speed. The conversation also includes a discussion on finding the velocity and position of the particle using integrals, as well as a suggestion to check for errors by pl
alex_b93

## Homework Statement

Particle (Mass = m) falling through a viscous liquid due to gravity.

Experiences a drag force that is proportional to its speed.
Magnitude of drag force .
F = ku = +ve Constant x velocity.

At time t = 0, z = position = 0, initial speed = u_0.

## Homework Equations

Acceleration found to be

du/dt = g - (k/m)u

## The Attempt at a Solution

So to find the velocity this needs to be integrated, the way I attempted to do it was to separate the variables and substitute in
α = mg / k
Meaning
(du/dt) = g(1-u/α)
This lead to the integral

∫(α du/ α-u) = ∫g dt
α∫(1/-u) du = g ∫dt

-α ln (α-u) = gt + c

ln (α-u) = -(gt + c )/α
(α-u) = exp -(gt + c)/α

u = α - exp[-(gt+c)/α]

Would that be correct up to there, because when I carry on to find c and to substitute for α it gets messy, so I'm unsure if I've done it right.

Thanks

Suggestion:

du/dt = g - (k/m)u

Above is correct equation. Solve the homogeneous equation by separation of variables. Then visualize a particular solution and add solutions together. The particular solution is 'particularly' easy to see. Evaluate constant by initial condition.

exp(ax+b)=d*exp(ax)

a,b,d are constants.

LawrenceC said:
Suggestion:

du/dt = g - (k/m)u

Above is correct equation. Solve the homogeneous equation by separation of variables. Then visualize a particular solution and add solutions together. The particular solution is 'particularly' easy to see. Evaluate constant by initial condition.

Okay so I ended up with

u(t) = [U_0 - (gm/k)] exp(-kt/m) + (gm/k)

Is that anywhere near correct?

That is what I got. You can 'test' your solution. Let t go to infinity and the velocity is mg/k which agrees with a force balance on the object. At t=0, the solution is V=V0.

Good work!

LawrenceC said:
That is what I got. You can 'test' your solution. Let t go to infinity and the velocity is mg/k which agrees with a force balance on the object. At t=0, the solution is V=V0.

Good work!

Awesome, thanks mate.

If I were to need to find the position I would need to integrate that, right?

Just that I need in a specific form, and I can't seem to get it in that form.

Yes, the integral of velocity over time provides displacement. Displacement is the area under the velocity-time curve. With an indefinite integral, you'll get another constant of integration that you evaluate from the initial condition at t=0 which the problem stated is z=0.

LawrenceC said:
Yes, the integral of velocity over time provides displacement. Displacement is the area under the velocity-time curve. With an indefinite integral, you'll get another constant of integration that you evaluate from the initial condition at t=0 which the problem stated is z=0.

Yeah okay, so this is what I ended up with as the position

z(t) = [( m exp(kt /m)(g m-k U_0)) /k + gmt) ]/k + c

Which is quite hard to understand when wrote on here.

But is that correct, trying to work out when I'm going wrong as I can't get it in the right form.
If that is right at least I know I should be able to write it in that form.

Cheers again.

You essentially have:

z(t) = integral[((Vo - mg/k)exp(-kt/m) + mg/k)*dt]

Vo-mg/k and mg/k are constants so you would get after integration

z(t) = (-m/k)[Vo - mg/k]exp(-kt/m) + mgt/k + C

with C the constant of integration that is evaluated when t=0.

Also, your parentheses do not balance.

Last edited:
When deriving formulas/functions, it is prudent to plug in the units to ensure you have not made a careless error. If the units are not correct, you have an error.

LawrenceC said:
You essentially have:

z(t) = integral[((Vo - mg/k)exp(-kt/m) + mg/k)*dt]

Vo-mg/k and mg/k are constants so you would get after integration

z(t) = (-m/k)[Vo - mg/k]exp(-kt/m) + mgt/k + C

with C the constant of integration that is evaluated when t=0.

Also, your parentheses do not balance.

Don't know how I missed the whole constant thing, feel a bit of an idiot now.

I'm getting

(V0 - mg/k)+ mg/k ∫exp(-kt/m)

[(V0 - mg/k)+ mg/k] [-m/k exp(-kt/m) + c]

Am I heading the right way now?

alex_b93 said:
Don't know how I missed the whole constant thing, feel a bit of an idiot now.

I'm getting

(V0 - mg/k)+ mg/k ∫exp(-kt/m)

[(V0 - mg/k)+ mg/k] [-m/k exp(-kt/m) + c]

Am I heading the right way now?

You still have an error. Write it like this and integrate:

z(t)= (V0 - mg/k)*integral(exp(-kt/m))dt + integral(mg/k)dt

LawrenceC said:
You still have an error. Write it like this and integrate:

z(t)= (V0 - mg/k)*integral(exp(-kt/m))dt + integral(mg/k)dt

Ah right, now I'm getting answers that match up now, I didn't think to do it that way, thank you for the advice.
Problem is now though the form it is wanted in is extremely similar to what I'm getting, or rather we're getting, except the last term is

[1-exp(-kt/m)]

Rather than

[-exp(-kt/m)]

Am I missing something really simple?

OK, when you integrate you pick up another constant of integration that must be evaluated. See if that helps.

LawrenceC said:
OK, when you integrate you pick up another constant of integration that must be evaluated. See if that helps.

Okay I got it, cheers for the help mate.

I have one last question to do, and would appreciate it if you could perhaps give me a nudge in the right direction with it.

It leads on from the last question.

Position of a particle acting under a constant force F in the same z direction is given by

z = u0 t + (F/2m)t^2

With the same initial conditions.
It asks me to show that the position of a particle falling under the influence of gravity through a viscous fluid can be approximated by that of a particle acted upon by a uniform force F, for sufficiently small values of t.
It then asks for an expression for F.

All I can think of is t≈0, and then to substitute this in.

alex_b93 said:
Okay I got it, cheers for the help mate.

I have one last question to do, and would appreciate it if you could perhaps give me a nudge in the right direction with it.

It leads on from the last question.

Position of a particle acting under a constant force F in the same z direction is given by

z = u0 t + (F/2m)t^2

With the same initial conditions.
It asks me to show that the position of a particle falling under the influence of gravity through a viscous fluid can be approximated by that of a particle acted upon by a uniform force F, for sufficiently small values of t.
It then asks for an expression for F.

All I can think of is t≈0, and then to substitute this in.

Talking to people in class it apparently uses the approximation for exp, I'm so confused.

## What is velocity and how is it related to acceleration?

Velocity is a measure of an object's speed and direction. It is related to acceleration because acceleration is the rate of change of an object's velocity. This means that an object's velocity increases or decreases by a certain amount over a specific period of time.

## How do you calculate velocity from acceleration?

To calculate velocity from acceleration, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. You can also use the formula v = d/t, where d is the distance traveled and t is the time taken. These formulas can be used for both linear and angular acceleration.

## What units are used to measure velocity and acceleration?

The SI unit for velocity is meters per second (m/s) and the SI unit for acceleration is meters per second squared (m/s²). However, other units such as kilometers per hour (km/h) and miles per hour (mph) are also commonly used to measure velocity. Similarly, units like feet per second squared (ft/s²) and miles per hour per second (mph/s) can be used to measure acceleration.

## Can velocity and acceleration be negative?

Yes, velocity and acceleration can both be negative. A negative velocity indicates that an object is moving in the opposite direction of its positive velocity. Similarly, a negative acceleration indicates that an object is slowing down in the opposite direction of its positive acceleration.

## What are some real-world applications of finding velocity from acceleration?

One common real-world application of finding velocity from acceleration is in the automotive industry, where engineers use acceleration data to design cars with better performance and safety features. Another application is in sports, where athletes use velocity and acceleration data to improve their performance and technique. In the field of physics, velocity and acceleration data are used to analyze the motion of objects and understand the laws of motion.

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