Page:Calculus Made Easy.pdf/82

$v=-3$; so that the body was moving in the direction opposite to its motion in the previous cases. As the acceleration is positive, however, we see that this velocity will decrease as time goes on, until it becomes zero, when $v=0$ or $0.4t-3=0$; or $t=7.5$ sec. After this, the velocity becomes positive; and $5$ seconds after the body started, $t=12.5$, and

$v=0.4\times 12.5-3=2{\text{ ft./sec}}$.

When $x=100$,

$100=0.2t^{2}-3t+10.4,\quad {\text{or }}\quad t^{2}-15t-448=0$,

and

$t=29.95;\ v=0.4\times 29.95-3=8.98{\text{ ft./sec}}$.

When $v$ is zero, $x=0.2\times {7.5^{2}}-3\times {7.5}+10.4=-0.85$, informing us that the body moves back to $0.85$ ft. beyond the point $O$ before it stops. Ten seconds later

$t=17.5$ and $x=0.2\times 17.5^{2}-3\times 17.5+10.4=19.15$.

The distance travelled $=.85+19.15=20.0$ and the average velocity is again $2$ ft./sec.

(4) Consider yet another problem of the same sort with $x=0.2t^{3}-3t^{2}+10.4$; $v=0.6t^{2}-6t$; $a=1.2t-6$ The acceleration is no more constant.

When $t=0$, $x=10.4$, $v=0$, $a=-6$. The body is at rest, but just ready to move with a negative acceleration, that is to gain a velocity towards the point $O$.

(5) If we have $x=0.2t^{3}-3t+10.4$, then $v=0.6t^{2}-3$, and $a=1.2t$.

When $t=0$, $x=10.4$; $v=-3$; $a=0$.

The body is moving towards the point $O$ with