Page:Calculus Made Easy.pdf/80

 Examples.

(1) A body moves so that the distance $$x$$ (in feet), which it travels from a certain point $$O$$, is given by the relation $$x=0.2t^2+10.4$$, where $$t$$ is the time in seconds elapsed since a certain instant. Find the velocity and acceleration $$5$$ seconds after the body began to move, and also find the corresponding values when the distance covered is $$100$$ feet. Find also the average velocity during the first $$10$$ seconds of its motion. (Suppose distances and motion to the right to be positive.) Now $x = 0.2t^2 + 10.4 $, $v=\dot{x}=\frac{dx}{dt}=0.4t$; and $a = \ddot{x} = \frac{d^2x}{dt^2} = 0.4 =$ constant. When $$t=0$$, $$x=10.4$$ and $$v=0$$. The body started from a point $$10.4$$ feet to the right of the point $$O$$; and the time was reckoned from the instant the body started.

When $$v = 0.4 \times 5 = 2 \text{ft./sec.}$$; $$a = 0.4\,\text{ft./sec}^2$$.

When $$100 = 0.2t^2 + 10.4$$ or $$t^2 = 448$$,

and $$t = 21.17 \text{sec.}$$; $$v = 0.4 \times 21.17 = 8.468\,\text{ft./sec.}$$

When $$t=10$$

distance travelled $$= 0.2 \times 10^2 + 10.4 - 10.4 = 20 \,\text{ft.}$$

Average velocity = $$\tfrac{20}{10} = 2 \,\text{ft./sec.}$$

(It is the same velocity as the velocity at the middle of the interval, $$t=5$$; for, the acceleration being constant, the velocity has varied uniformly from zero when $$t=0$$ to $$4\,\text{ft./sec.}$$. when $$t=10$$.)