Physics Lab » Equations of Motion

Jul 28 2008

Equations of Motion

Equations of motion are equations that describe the behaviour of a system or particle under the influence of a force. The equations are usually given as a function of time but equations as a function of velocity or energy are also given.

In classical Newtonian Mechanics, equations of motion usually describe the position or the velocity of a particle as a function of time; such as an object falling under the influence of gravity. In this section we will examine motion in one dimension of an object or particle under the influence of a constant force.

Acceleration Under Constant Force

We first start with Newton’s Second Law; which states that the acceleration of a particle is proportional to the net force applied on the particle and inversely proportional to the mass of the particle. This is usually the first place we start when solving equations of motion:

$F=ma$

(1)

The acceleration of a particle is defined as the change in particle’s velocity with respect to time, or in the differential case, a = dv/dt.

$F=m\frac{dv}{dt}$

(2)

Let us consider the simplest case of a particle under the influence of a constant force. F = F0. So the equation we have to solve is simply:

$F_0=m\frac{dv}{dt}$

(3)

Equation 3 is a separable differential equation and is easy to solve. Separating the variables we have:

$dv = \frac{F_0}{m}dt$

(4)

We now integrate both sides.

$\int dv = \int \frac{F_0}{m}dt$

$v = \frac{F_0}{m}t +C$

(5)

By inspection we see that if t=0, then v = C, or C is the initial velocity of the particle. Going back to equation 3, we see that a = dv/dt = F0/m, so we can can write equation 5 as:

$v(t) = at + v_0$

(6)

To determine the position as a function of time, we must solve another equation of motion. Velocity is defined as the change in position with respect to time, in the differential case, v = dx/dt:

$\frac{dx}{dt}= at+v_0$

(7)

Separating the variables again and integrating we get:

$dx=(at+v_0)dt$

$\int dx = \int (at+v_0)dt$

$x(t)=\frac{1}{2}at^2+v_0t+D$

(8)

Again, by inspection, if we let t=0, we see that x(0) = D, so D is just the initial position of the particle with respect to some origin.

$x(t)=\frac{1}{2}at^2+v_0t+x_0$

(9)

So we see that the motion of the particle under the influence of a constant force is given by the following two equations, the position as a function of time (equation 9), and the velocity as a function of time (equation 6):

Retarding Forces (Drag)

A retarding force is a force that tends to retard or restrict the motion of a particle. The most common being drag or air friction. For velocities under 25 m/s, the drag force is proportional to the velocity of the object, but in the opposite direction. For higher velocities, the drag force is proportional to the square of the velocity. We will only consider slow velocities. The above applet assumes this as well.

$F_D = -kv$

(10)

The constant k can be related to other factors as well, such as air density, cross sectional area of the object, and mass. But we will combine all these factors into one constant, k, to make the math slightly simpiler.

The total force acting on the particle is the sum of all other forces; $F_0$ , our constant force applied to the object and $F_D$ , the drag force.

$F_T = F_0+F_D$

(11)

$F_T = F_0-kv$

(12

Using Newton’s law we get the differential equation of motion we must solve.

$F=ma$

(13)

$m\frac{dv}{dt}=F_0-kv$

(14)

Separating the variables and integrating we get:

$m\frac{dv}{F_0-kv}=dt$

(15)

$m\int \frac{dv}{F_0-kv}=\int dt$

(16)

$-\frac{m}{k}\ln|F_0-kv|=t+E$

(17)

Solving for v, we get:

$v(t)=\frac{1}{k}( F_0 + Ge^{-\frac{kt}{m}})$

(18

Here, E is the constant of integration and G is another constant after various operations on E. i.e: a constant times a multiple is also a constant, the sine of a constant is another constant.

Given our initial condition $v(0) = v_0$ , we substitute $t=0$ and $v(t)=v_0$ . We can solve for G:

$G = v_0k-F_0$

(19)

Substituting equation 19 into equation 18 and rearranging we get our equation of velocity.

$v(t)=\frac{F_0}{k} + (v_0 - \frac{F_0}{k})e^{ -\frac{kt}{m} }$

(20)

Notice that:

$\mathop{\lim}\limits_{t \to \infty} v(t) = \frac{F_0}{k}$

(21)

As time increases, the velocity reaches a stable value, this is the terminal veclotiy of the particle. As the particle accelerates and gains speed, the drag force builds up and eventually balances out the applied force. At this time the particle stops accelerating and moves with a constant speed given by equation 21.

To determine the position as a function of time, we integrate the velocity function and use the initial conditions x(0) = x0 to determine the constant of integration. Doing this yeilds the postion function:

$x(t) = \frac{F_0}{k}t - \frac{m}{k}( v_0 - \frac{F_0}{k})e^{ -\frac{kt}{m}} + \frac{m}{k}(v_0 - \frac{F_0}{k})+x_0$

(22)

Units of k

If we take a look at equation 21, we see that on the left side of the equation we have a velocity parameter with units m/s. On the right side of the equation we have a force with units N (newtons) divided by the constant k.

To be a valid physics equation that relates one property to another, the units must be the same. Therefore, k has some units that makes the right side of the equation have the same units as the left side.

The units of force is Newtons, but in the base units, it is:

$N \equiv kg\frac{m}{s^2}$

(23)

If we multiply the left and right side of equation 21 by k, we get a rearranged formula for k.

$k \equiv \frac{F_0}{v}$

(24)

Now to determine the units for k, we simply input in the units for force and velocity into the right side and see what we get.

$k \equiv kg\frac{m}{s^2}\frac{s}{m}$

(25)

We see that metres and one of the time units can divide out, and we are left with.

$k \equiv \frac{kg}{s}$

(26)

So the units for the drag constant is $\frac{kg}{s}$ .

Have any questions about this information? Please contact me.

7 responses so far

7 Responses to “Equations of Motion”

# Matt Gilgenbachon 10 Oct 2008 at 1:01 am

Equation 22 has some typos. It should read x(t) = F0/k*t – m/k(v0-F0/k)e^(-k*t/m)+ m/k*(v0 – F0/k) + x0

The k/m coefficients on the second and third terms should be m/k.

Otherwise, the page is very helpful!

-Matt
# adminon 10 Oct 2008 at 10:00 am

The typos have been fixed. Thank you for noticing that, Matt.
# frankon 06 Nov 2008 at 2:47 pm

There is another typo between eqn (7) and (8) where you meant to write
dx = (at + vo)dt and you wrote dx=(at+vo)t

Very nice site and a great reference! I had a hard time finding sites that did a classic derivation like this
# John PlurnDoubberoon 18 Dec 2008 at 8:17 am

First of all congratulation for such a great site. I learned a lot reading article here today. I will make sure i visit this site once a day so i can learn more.
# tobyon 24 Feb 2009 at 11:06 am

another typo is in equation 16:

You have in the denominator: F_0 – kdv
it should be: F_0 – kv

so instead of dv it should read a simple v.
# adminon 26 Feb 2009 at 6:24 pm

Thank you Toby, I have fixed the typo.
# lachlanon 15 Sep 2009 at 2:27 am

if you could put different levels of difficulty in the equations like having the simplest forms for;
force =
velocity =
acceleration =

cause not everyones a jenis lol

Equations of Motion

Equations of Motion

Acceleration Under Constant Force

Retarding Forces (Drag)

Units of k

7 Responses to “Equations of Motion”

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