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 $a={\ddot {x}}={\frac {d^{2}x}{dt^{2}}}=0.4=$ 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 $t=10$.)