Page:Calculus Made Easy.pdf/81

(2) In the above problem let us suppose

$x=0.2t^{2}+3t+10.4$.

$v={\dot {x}}={\dfrac {dx}{dt}}=0.4t+3;\quad a={\ddot {x}}={\frac {d^{2}x}{dt^{2}}}=0.4={\text{constant}}$.

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

When $x=100$ ft.,

$100=0.2t^{2}+3t+10.4$; or $t^{2}+15t-448=0$;

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

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

When $t=-7.5$,

$x=0.2\times {(-7.5)^{2}}-3\times {7.5}+10.4=-0.85\,{\text{ft}}.$,

that is $0.85$ ft. to the left of the point $O$.

Now, when $t=2.5$,

$x=0.2\times {2.5^{2}}+3\times {2.5}+10.4=19.15$.

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

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