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^2x}{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