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.(a) How far does the ball travel during the time interval [4, 4.5] ?____ ft(b) Compute the average velocity over (4, 4.5) .____ft/sec(c) Compute the average velocity over time intervals [4, 4.01] , [4, 4.001] , [4, 4.0001] , [3.9999, 4] , [3.999, 4] , [3.99, 4]Use this to estimate the object's instantaneous velocity at t=4V(4) = ____ft/sec

Answer 1

Given that the distance traveled by the ball after t seconds is

`s(t)=16t^2\text{ ft}`

(a) We have to find the distance traveled by the ball during the time interval [4,4.5]

Here, the time t1 = 4 sec

The time t2 = 4.5 sec

The distance traveled during this time interval will be

`\begin{gathered} s(t2)-s(t1)=16(4.5)^2-16(4)^2 \\ =324-256 \\ =\text{ 68 ft} \end{gathered}`

Thus, the ball traveled 68 ft during the time interval [4,4.5].

(b) We have to calculate the average velocity over the time t1 = 4 sec and

t2 = 4.5 sec.

The average velocity will be

`\begin{gathered} v_(av)=(16*(4.5)^2-16*(4)^2)/(4.5-4) \\ =136\text{ ft/sec} \end{gathered}`

Thus the average velocity is 136 ft/sec over time (4,4.5)

(c) We have to find the average velocity over the time intervals

[4.4.01], [4, 4.001],[4,4.0001],[3.9999,4],[3.999,4],[3.99,4]

The average velocity for the time interval [4,4.01] is

`\begin{gathered} v_(av)1\text{ = }\frac{16*(4.01)^2^{}-16*(4)^2}{4.01-4} \\ =128.16\text{ ft/sec} \end{gathered}`

The average velocity for the time interval [4,4.001] is

`\begin{gathered} v_(av)2\text{ = }(16*(4.001)^2-16*(4)^2)/(4.001-4) \\ =128.016\text{ ft/sec} \end{gathered}`

The average velocity for the time interval [4,4.0001] is

`\begin{gathered} v_(av)3\text{ = }(16*(4.0001)^2-16*(4)^2)/(4.0001-4) \\ =128.0016\text{ ft/sec} \end{gathered}`

The average velocity for the time interval [3.9999,4] is

`\begin{gathered} v_(av)4\text{ = }(16*(4)^2-16*(3.9999)^2)/(4-3.9999) \\ =127.9984\text{ ft/sec} \end{gathered}`

The average velocity for the time interval [3.999,4] is

`\begin{gathered} v_(av)5\text{ = }(16*(4)^2-16*(3.999)^2)/(4-3.999) \\ =127.984\text{ ft/sec} \end{gathered}`

The average velocity for the time interval [3.99,4] is

`\begin{gathered} v_(av)6\text{ = }(16*(4)^2-16*(3.99)^2)/(4-3.99) \\ =127.84\text{ ft/sec} \end{gathered}`

We can estimate the object's instantaneous velocity at t = 4 sec as

V(4) = 128 ft/sec

As all the values are nearing to 128 ft/sec