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 $$\dfrac{d^2y}{dt^2}$$ is large. When it has got up its top speed it is no longer being accelerated, so that then $$\dfrac{d^2y}{dt^2}$$ 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 $$\dfrac{d^2y}{dt^2}$$ 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 $f=ma$;