Answer
a. $\frac{4}{7}$ sec. $280cm/s$
b. $560cm/s$, $980cm/s^2$
c. $30$ flashes/sec.
Work Step by Step
a. Given $s(t)=490t^2$, we can find the velocity as $v(t)=s'(t)=980t$ ft/s and $a=980cm/s^2$. Letting $s(t)=160$, we have $490t^2=190$ which gives $t=\sqrt {16/49}=\frac{4}{7}$ sec. The average velocity can be found as $\bar v=\Delta s/\Delta t=\frac{160}{(\frac{4}{7})}=280cm/s$
b. At $t=\frac{4}{7}$ sec, we have $v=980(\frac{4}{7})=560cm/s$, and $a=980cm/s^2$
c. We can count 17 photos from the start to the time when the ball reached $160cm$; the time interval was $\frac{4}{7}$ sec, thus we have $n=17/(\frac{4}{7})\approx30$ flashes/sec.
