Page:Calculus Made Easy.pdf/81

(2) In the above problem let us suppose

When ${\displaystyle t=0}$, ${\displaystyle x=10.4}$ and ${\displaystyle v=3}$ ft./sec, the time is reckoned from the instant at which the body passed a point ${\displaystyle 10.4}$ ft. from the point ${\displaystyle O}$, its velocity being then already ${\displaystyle 3}$ ft./sec. To find the time elapsed since it began moving, let ${\displaystyle v=0}$; then ${\displaystyle 0.4t+3=0}$, ${\displaystyle t=-{\tfrac {3}{.4}}=-7.5}$ sec. The body began moving ${\displaystyle 7.5}$ sec. before time was begun to be observed; ${\displaystyle 5}$ seconds after this gives ${\displaystyle t=-2.5}$ and ${\displaystyle v=0.4\times -2.5+3=2}$ ft./sec.

When ${\displaystyle x=100}$ ft.,

hence ${\displaystyle t=14.95}$ sec., ${\displaystyle v=0.4\times {14.95}+3=8.98}$ ft./sec.

To find the distance travelled during the ${\displaystyle 10}$ first seconds of the motion one must know how far the body was from the point ${\displaystyle O}$ when it started.

When ${\displaystyle t=-7.5}$,

that is ${\displaystyle 0.85}$ ft. to the left of the point ${\displaystyle O}$.

Now, when ${\displaystyle t=2.5}$,

So, in ${\displaystyle 10}$ seconds, the distance travelled was ${\displaystyle 19.15+0.85=20}$ ft., and the average velocity ${\displaystyle ={\tfrac {20}{10}}=2{\text{ ft./sec}}}$.

(3) Consider a similar problem when the distance is given by ${\displaystyle x=0.2t^{2}-3t+10.4}$. Then ${\displaystyle v=0.4t-3}$, ${\displaystyle a=0.4=}$ constant. When ${\displaystyle t=0}$, ${\displaystyle x=10.4}$ as before, and