Linear Velocity and Acceleration

Key facts

Velocity is the rate of change of position with respect to time:

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

(1)

Acceleration is the rate of change of velocity with respect to time:

$a = \frac{dv}{dt} (=\dot v)$

(2)

$a = \frac{d^2s}{dt^2} (=\ddot s)$

(3)

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

(4)

If $u$ is the initial velocity, $v$ the final velocity, and $s$ the distance traveled in a time $t$ , then:

$v = u + at$

(5)

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

(6)

and

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

(7)

Velocity and acceleration are vector quantities both measured with respect to time. While velocity is the rate of change of position of an object, acceleration is the rate of change of velocity.

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Introduction

Velocity and Acceleration are terms used to describe the state of motion of an object.

The term velocity identifies the speed and direction in which the object is moving. Imagine you were sitting in a bus that is travelling on a long stretch of road between two towns A and B. If the towns are separated by a certain known distance, and you can measure the time it took you to reach the other town, you can then tell the velocity at which your bus was travelling. Also, if the same bus was travelling between the two towns A and B at a constant speed, we would say that the bus was travelling at Constant Linear Velocity.

What would happen, however, if the bus happened to change its velocity while travelling between the two towns, that is, by speeding up or slowing down? We would then say that during the time that the bus was changing its velocity (speeding up or slowing down) it was in a state of acceleration.

Acceleration therefore, is the rate at which the object changes its velocity. Constant acceleration, the term in consideration here, assumes that the rate of change of velocity over a period of time is constant. In other words, if the bus was in a state of constant acceleration, we would say that the bus was either speeding up or slowing down at a steady, constant rate.

Linear Velocity

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. Therefore, notations such as $\displaystyle \frac{dv}{dt}$ , $\displaystyle \frac{ds}{dt}$ , $\displaystyle \frac{d^2s}{dt^2}$ , 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 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 $600+29.33t ft$ :