Page:Calculus Made Easy.pdf/75

the distance passed over and the time taken to pass over it. You are dividing one by the other. If $y$ is the whole distance, and $t$ the whole time, clearly the average rate is ${\dfrac {y}{t}}$. Now the speed was not actually constant all the way: at starting, and during the slowing up at the end of the journey, the speed was less. Probably at some part, when running downhill, the speed was over $60$ miles an hour. If, during any particular element of time $dt$, the corresponding element of distance passed over was $dy$, then at that part of the journey the speed was ${\dfrac {dy}{dt}}$. The rate at which one quantity (in the present instance, distance) is changing in relation to the other quantity (in this case, time) is properly expressed, then, by stating the differential coefficient of one with respect to the other. A velocity, scientifically expressed, is the rate at which a very small distance in any given direction is being passed over; and may therefore be written

$v={\dfrac {dy}{dt}}$.

But if the velocity $v$ is not uniform, then it must be either increasing or else decreasing. The rate at which a velocity is increasing is called the acceleration. If a moving body is, at any particular instant, gaining an additional velocity $dv$ in an element of time $dt$, then the acceleration $a$ at that instant may be written

$a={\dfrac {dv}{dt}}$;