ANSWERS
(a) 68 ft
(b) 136 ft/s
(c) 128 ft/s
EXPLANATION
(a) The time interval is from 4s to 4.5s, so the distance the ball travels from 4s to 4.5s is,
[tex]\Delta s=16\cdot(4.5)^2-16(4)^2=68ft[/tex]
(b) As stated, the average velocity is the quotient between the distance traveled and the time,
[tex]\frac{\Delta s}{\Delta t}=\frac{68ft}{0.5s}=136ft/s[/tex]
(c) Here we have to find the distance as we did in part b and then divide by the time interval,
[tex]\begin{cases}\lbrack4,4.01\rbrack\to\Delta s=1.28016\to V=1.28016/0.01=128.16ft/s \\ \lbrack4,4.001\rbrack\to\Delta s=0.128016\to V=0.128016/0.001=128.016ft/s \\ \lbrack4,4.0001\rbrack\to\Delta s=0.01280016\to V=0.01280016/0.0001=128.0016ft/s \\ \lbrack3.9999,4\rbrack\to\Delta s=0.01279984\to V=0.01279984/0.0001=127.9984ft/s \\ \lbrack3.999,4\rbrack\to\Delta s=0.127984\to V=0.127984/0.001=127.984ft/s \\ \lbrack3.99,4\rbrack\to\Delta s=1.2784\to V=1.2784/0.01=127.84ft/s\end{cases}[/tex]
As we can see in the middle values, as the time interval is shorter - the difference approaches 0, the value of the velocity is closer to 128ft/s.
Hence, the estimated instantaneous velocity at t = 4 is 128 ft/s
