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An introduction to linear velocity and acceleration, also taking into account the particular case of constant acceleration

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Introduction

In order to define linear velocity, consider a particle moving in a straight line from a fixed point $O$ , as diagramed in Figure 1.

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 $\delta\,t$ the particle traveled the distance $\delta\,s$ , from which it can be seen that the average rate of displacement from $O$ , or average velocity, is $\displaystyle \frac{\delta\,s}{\delta\,t}$ . The velocity at time $t$ is defined as the limiting value of this quantity as $\delta\,t\rightarrow 0$ :

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

(8)

Thus, velocity can be defined as the rate of change of position with respect to time.

Similarly, it can be shown that acceleration, which is the rate of change of velocity with respect to time, is $\displaystyle\frac{dv}{dt}$ , or, expressed in a different form, $\displaystyle\frac{d^2s}{dt^2}$ . Acceleration can also be written in a third form which is independent of time. For example, by expressing the acceleration as:

$\frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt}$

and taking into account (#1), we get:

$\frac{dv}{dt} = v \frac{dv}{ds}$

Hence, acceleration may be expressed in any of the following three forms:

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

It is usual in Mechanics to denote differential coefficients with respect to time by dots placed above the dependent variables, so that notations as for example $\displaystyle \frac{dv}{dt}$ , $\displaystyle \frac{ds}{dt}$ , $\displaystyle \frac{d^2s}{dt^2}$ , and $\displaystyle\frac{d^2x}{dt^2}$ can also be denoted by $\dot{v}$ , $\dot{s}$ , $\ddot{s}$ , and $\ddot{x}$ respectively.

Worked example #564 not found.

Constant Acceleration

Let $u$ be the initial velocity, $v$ the final velocity, $a$ the acceleration, which in this case is a constant, $s$ the distance traveled in a time $t$ , and $t$ the time.

We previously saw that the acceleration can be written as $\displaystyle \frac{dv}{dt}$ . As in this case $a$ is constant, by integrating $a = \displaystyle \frac{dv}{dt}$ with respect to $t$ , we get:

$v = at + Const.$

(9)

In this equation at $t = 0$ , $v$ becomes $u$ , and so the constant equals $u$ . Thus, (#2) becomes:

$v = u + at$

Integrating again with respect to $t$ , and considering that $s = \displaystyle\frac{ds}{dt}$ , we get:

$s = ut + \frac{1}{2} a t^2 + Const.$

(10)

In this equation at $t = 0$ , $s$ becomes $0$ , and thus the constant equals $0$ . Hence, (#3) becomes:

$s = ut + \frac{1}{2} a t^2$

Integrating this equation with respect to $s$ , and considering that $v \displaystyle\frac{dv}{ds} = a$ , we get:

$\frac{1}{2} v^2 = as + Const.$

(11)

As at $s = 0$ , $v$ becomes $u$ , the constant becomes $\displaystyle \frac{1}{2} u^2$ . Thus (#4) becomes:

$v^2 = u^2 + 2as$

\calc{(v^2-u^2)/2s} "Instant calculator eq(10)"

Example 1 [imperial]

Problem

The driver of an express train traveling at 60 mph sees, on the same track, 600 ft in front of him, a slow train traveling in the same direction at 20 mph.

What is the least retardation that must be applied to the express to avoid a collision?

Workings

For the express train, $u = 88ft/s$ , and $v = 29.33ft/s$ . Taking into account that $v^2=u^2+2as$ , we get:

$29.33^2 - 88^2 = 2as$

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