Page:Calculus Made Easy.pdf/83

a velocity of $3$ ft./sec., and just at that instant the velocity is uniform.

We see that the conditions of the motion can always be at once ascertained from the time-distance equation and its first and second derived functions. In the last two cases the mean velocity during the first $10$ seconds and the velocity $5$ seconds after the start will no more be the same, because the velocity is not increasing uniformly, the acceleration being no longer constant.

(6) The angle $\theta$ (in radians) turned through by a wheel is given by $\theta ==3+2t-0.1t^{3}$, where $t$ is the time in seconds from a certain instant; find the angular velocity $\omega$ and the angular acceleration $\alpha$, (a) after $1$ second; (b) after it has performed one revolution. At what time is it at rest, and how many revolutions has it performed up to that instant?

Writing for the acceleration

$\omega ={\dot {\theta }}={\dfrac {d\theta }{dt}}=2-0.3t^{2},\quad \alpha ={\ddot {\theta }}={\dfrac {d^{2}\theta }{dt^{2}}}=-0.6t$.

When $t=0$, $\theta =3$; $\omega =2$ rad./sec.; $\alpha =0$.

When $t=1$,

$\omega =2-0.3=1.7\,{\text{rad./sec.}};\quad \alpha =-0.6\,{\text{rad./sec}}^{2}$.

This is a retardation; the wheel is slowing down.

After $1$ revolution

$\theta =2\pi =6.28;\quad 6.28=3+2t-0.1t^{3}$.

By plotting the graph, $\theta =3+2t-0.1t^{3}$, we can get the value or values of $t$ for which $\theta =6.28$; these are $2.11$ and $3.03$ (there is a third negative value).