Answer
a. $-8.875ft/sec$
b. $0.123 rad/sec$. $-0.123 rad/sec$
c. $\frac{1}{6} rad/sec$, $-\frac{1}{6} rad/sec$
Work Step by Step
a. The player’s distance from third base is given by $s^2=90^2+(90-x)^2$, where $x$ is the distance from the first base. We have $2ss'=-2(90-x)x'$. At $x=30, s=\sqrt {90^2+60^2}\approx108.2$, thus $108.2s'=-(90-30)(16)$ which gives $\frac{ds}{dt}=-8.875ft/sec$
b. $tan\theta_2=\frac{90-x}{90}$ and $sec^2\theta_2\frac{d\theta_2}{dt}=-\frac{1}{90}\frac{dx}{dt}=-\frac{16}{90}$. With $cos\theta_2=\frac{90}{108.2}$, we have $\frac{d\theta_2}{dt}=-\frac{16}{90}(\frac{90}{108.2})^2\approx-0.123 rad/sec$. Similarly, as $\theta_1=\pi/2-theta_2$, we have $\frac{d\theta_1}{dt}=-\frac{d\theta_2}{dt}=0.123 rad/sec$
c. For this case, $x=90, s=90, \frac{dx}{dt}=15ft/sec, sec^2(0)=1$, we have $(1)\frac{d\theta_2}{dt}=-\frac{1}{90}(15)$, which gives $\frac{d\theta_2}{dt}=-\frac{1}{6} rad/sec$ and $\frac{d\theta_1}{dt}=\frac{1}{6} rad/sec$
