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Applied mathematics

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Linear Velocity and Acceleration

Velocity is defined as the rate of change of position. For linear velocity consider a particle moving in a straight line either to or from a fixed point O. At a time t let the distance from O be s and at a time $t\:+\:\delta\,t$ let the distance increase to $s\:+\:\delta\,s$ . So over a time of $\delta\,t$ has traveled $\delta\,s$ from which it can be seen that the average rate of displacement from O or average velocity is $\frac{\delta\,s}{\delta\,t}$

The velocity at time t is defined as the limiting value of this quantity as $\delta\,t\;\rightarrow\:0$

$\therefore\;\;\;v\;=\;\frac{ds}{dt}$

(1)

Similarly it can be shown that acceleration (the rate of change of velocity with respect to time) is $\frac{dv}{dt}$ which can also be expressed as $\frac{d^2s}{dt^2}$ .

Acceleration can also be written in a third form which is independent of the time. Velocity is a function of the distance s

$\therefore\:\:\;\frac{dv}{dt}\;=\;\frac{dv}{ds}\:X\:\frac{ds}{dt}\;which \;=\;{v\:\frac{dv}{ds}$

(2)

Hence acceleration may be expressed in any one of these three forms

$\frac{dv}{dt};\;\:\frac{d^2s}{dt^2};\:\texttt{or}\;v\:\frac{dv}{ds}$

(3)

It is usual in Mechanics to denote differential coefficients with respect to time by dots placed above the dependent variables so that

$\frac{dv}{dt}\,;\:\frac{ds}{dt}\,;\:\frac{d^2s}{dt^2}\,and\:\frac{d^2x}{dt^2}\:\texttt{are denenoted by}\:\dot{v}\,;\:\dot{s}\,;\:\ddot{s}\:and\:\ddot{x}\:\texttt{respectively}$

(4)

Constant Acceleration

Let

u = the initial velocity
v = the final velocity
a = acceleration which in case is a constant
s = the distance traveled in a time t
t = time

We know that $\frac{dv}{dt}$ = acceleration, which in this case is a constant a Integrating

$v\;=\;at\;+\;Const.$

(5)

At t=0, v=u and so C=u

$\therefore\;\;\;v\;=\;u\;+\;a\,t$

(6)

Integrating again since $s\;=\;\frac{ds}{dt}$

$s\:=\;u\,t\:+\;\frac{1}{2}\:a\,t^2\;+\;Const$

(7)

But when t=0 s=0 and so the constant equals 0 so

$s\;=\;u\,t\;+\;\frac{1}{2}\:a\,t^2$

(8)

We can also write $v\:\frac{dv}{ds}\;=\;a$

Integrating with respect to s

$\frac{1}{2}\:v^2\;=\;a\:s\;+\;Const,$

(9)

But when s=0, v=u and so $C\;=\;\frac{1}{2}\:u^2$ so

$v^2\;=\;u^2\:+\:2\,a\,s$

(10)

Example 1

The driver of an express train traveling at 60 m.p.h. sees, on the same track, 600 ft in front of him, a slow train traveling in the same direction at 20 m.p.h. What is the least retardation that must be applied to the express to avoid a collision?

For the express U = 88 ft/sec v = 29.33 ft/sec

Substituting in equation 23 above

$(29.33)^2\;-\;(88.0)^2\;=\;2\,a\,s$

(11)

In a time t the slow train will have traveled a distance 29.33t ft. The express will have gone further and will have traveled 600 + 29.33tft

$\therefore \;\:\;(29,33)^2\:-\:(88)^2\:=\:2\,a\,(600\:+\:29.33t)$

(12)

However from equation 16

$\frac{29.33\:-\:88}{a}\;=\;t$

(13)

Combining equations 25 and 26

$29.33^2\:-\:88^2\;=\;2\,a(600\:+\:29,33\left(\frac{29.33\,-\,88}{a} \right)$

(14)

From which

$a\;=\;-\:2.87\,ft./sec^2$

(15)

Further examples of constant acceleration can be seen in "Frictionless projectiles"

Simple Harmonic Motion

Let s be given by the equation $s\;=\:m\,cos\:n\,t$ where m and n are constants (miss the value of s obtained by putting t=0. i.e.it is the initial distance from the origin)

Then the velocity

$v\;=\:\frac{ds}{dt}\;=\;-\:mn\:sin\:nt$

(16)

and the acceleration

$=\;\frac{dv}{dt}\;=\;-\:m\,n^2\:cos\:nt\;=\:-n^2\,s$

(17)

Again eliminating t from the above two equations

$v^2\:=\:m^2\,n^2\:sin^2\:n\,t\;=\:n^2(m^2\:-\:m^2\,cos^2\,n\,t)\;=\:n^2(m^2\:-\:s^2)$

(18)

which gives v in terms of s Differentiating this in respect to s

$2\,v\,\frac{dv}{ds}\;=\;n^2(-2\,s)$

(19)

i.e. The acceleration is $-\;n^2\:s$ and is towards the origin and varies as the distance from the origin. This is simple harmonic motion ( See "Simple pendulum")