integrate

[Shay10825]

How would you integrate to find the velocity, displacement, and distance traveled :confused: ? Can someone give a quick example please.
~Thanks

[cepheid]

Say the position of a particle is described using a position vector \inline{ \vec{r}(t) = (x(t), y(t), z(t)) }. Notice that the position of the particle is a function of time (the three components vary with time).

I'm sure you know that velocity is the derivative of position with respect to time. In other words, velocity is defined as the rate of change of position with time or the rate of displacement with time. It need not be constant. The velocity too, can be a function of time.

\vec{v}(t) = \frac{d\vec{r}}{dt}

Similarly, acceleration is the rate of change of velocity with time:

\vec{a}(t) = \frac{d\vec{v}}{dt}

You asked how one would determine velocity and then displacement of a particle by integrating. The question only makes sense if you already know the acceleration of the particle as a function of time.

To find the velocity => you know its derivative (a) already. Just integrate it to find the original function. If you're not sure why, review your basic calculus: indefinite integrals are antiderivatives.

\frac{d\vec{v}}{dt} = \vec{a}(t)

d\vec{v} = \vec{a}(t)dt

\vec{v}(t) = \int{d\vec{v}} = \int{\vec{a}(t)dt}

Similarly:

\vec{r}(t) = \int{d\vec{r}} = \int{\vec{v}(t)dt}

That's the best way I could think of to explain it. This is really basic kinematics. Lots of problems involving specific functions are kicking around, in both your first-year calculus and first-year physics texts.

[maverick280857]

You need to know

(a) the integrand function (can be a function of time, displacement, velocity or even acceleration)

(b) the boundary values (or initial values) for the integral (usually the value at t = 0)

(c) the relationships that cepheid has mentioned

In addition, you have to come to terms with velocities not only as functions of time but also displacements and accelerations. The familiar equations for uniformly accelerated motion,

v(t) = v_{0}t + at
v^{2} = v_{0}^{2} + 2as
s = s_{0} + v_{0}t + \frac{1}{2}at^{2}

can actually be derived using integration (with the acceleration \doubledot x = a as a starting point). Try doing that to get a feel of calculus applied to kinematics.

Then, there's also a useful identity, a = v\frac{dv}{ds} that you could use (prove it yourself to understand it).

Essentially, derivatives and integrals are complementary. You will encounter them virtually throughout an undergraduate physics course, especially in dynamics. So you must get familiar with them to be able to convincingly use them in problems and situations. One of the main things is to realize when you have to resort to calculus. I believe you can understand that through the expressions cepheid has given in his post.

Enjoy physics,

Cheers
Vivek

[Gaz031]

Briefly:
If you have the displacement as a function of time you can differentiate to get the velocity. You can differentiate the velocity to get acceleration.
Similarly, you can integrate the acceleration to the velocity to the displacement. However, you need to add your constants carefully. If you have an expression for the acceleration you need to add the correct constant to the velocity expression in order to get the correct displacement expression. A lot of questions will give you information such as 'at time t=5, the displacement is 3 and the velocity is 6m/s'., this enables you to find the correct constants.

[Shay10825]

Similarly, you can integrate the acceleration to the velocity to the displacement.

Can you (or anyone else) give an example of how you would do this.

~Thanks

[maverick280857]

Supposing the acceleration function is given by

a = a(t)

Integrating once, I get the velocity as a function of time

v(t) = v_{0} + \int_{t = 0}}^{t}a(t) dt

Integrating the velocity, I get the displacement as a function of time

s(t) = s_{0} + \int_{t = 0}^{t}v(t) dt

As an elementary example, suppose a = g = constant. Integrating once, I get the velocity as

v = v_{0} + \int_{t = 0}^{t}g dt = v_{0} + gt

Integrating once more from t = 0 to t = t, I get

s = s_{0} + v_{0}t + \frac{1}{2}gt^{2}

where the subscript 0 indicates the value at t = 0.

Do these integrations yourself to convince yourself (I have intentionally left out some steps). I do think that you should refer to a freshman calculus text or a book like Thomas and Finney to get familiar with integration.