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).