Question

A ball is dropped from a state of rest at time T = 0.The distance traveled after t seconds is s(t) = 16t^2 ft.

Answer 1

ANSWERS

(a) 68 ft

(b) 136 ft/s

(c) 128 ft/s

Step-by-step explanation

(a) The time interval is from 4s to 4.5s, so the distance the ball travels from 4s to 4.5s is,

`\Delta s=16\cdot(4.5)^2-16(4)^2=68ft`

(b) As stated, the average velocity is the quotient between the distance traveled and the time,

`(\Delta s)/(\Delta t)=(68ft)/(0.5s)=136ft/s`

(c) Here we have to find the distance as we did in part b and then divide by the time interval,

`\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}`

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