Page:Calculus Made Easy.pdf/76

but $dv$ is itself $d\left({\dfrac {dy}{dt}}\right)$. Hence we may put $a={\frac {d\left({\tfrac {dy}{dt}}\right)}{dt}}$; and this is usually written $a={\frac {d\left({\tfrac {dy}{dt}}\right)}{dt}}$;

or the acceleration is the second differential coefficient of the distance, with respect to time. Acceleration is expressed as a change of velocity in unit time, for instance, as being so many feet per second per second; the notation used being $feet\div second^{2}$.

When a railway train has just begun to move, its velocity v is small; but it is rapidly gaining speed–it is being hurried up, or accelerated, by the effort of the engine. So its ${\dfrac {d^{2}y}{dt^{2}}}$ is large. When it has got up its top speed it is no longer being accelerated, so that then ${\dfrac {d^{2}y}{dt^{2}}}$ has fallen to zero. But when it nears its stopping place its speed begins to slow down; may, indeed, slow down very quickly if the brakes are put on, and during this period of deceleration or slackening of pace, the value of ${\dfrac {dv}{dt}}$, that is, of ${\dfrac {d^{2}y}{dt^{2}}}$ will be negative.

To accelerate a mass $m$ requires the continuous application of force. The force necessary to accelerate a mass is proportional to the mass, and it is also proportional to the acceleration which is being imparted. Hence we may write for the force $f$, the expression

$f=ma$;