Question

If an object is thrown upward at 128 feet per second from a height of 36 feet, its height S after t seconds is given by the following equation.S(t) = 36 + 128t â 16t2(a) What is the average velocity in the first 4 seconds after it is thrown?(b) What is the average velocity in the next 4 seconds?

Answer 1

(a) 64 feet per second.

(b) -64 feet per second.

Explanation:

We have been given that the height S after t seconds is given by the equation
`S(t) = 36+128t-16t^2`.

(a) We are find the average velocity in the first 4 seconds after it is thrown.

We will average rate change formula to solve our given problem.

`\text{Average rate change}=(f(b)-f(a))/(b-a)`

`\text{Average velocity in first 4 seconds}=(S(4)-S(0))/(4-0)`

Let us find S(4) and S(0) as:

`S(4)=36+128(4)-16(4)^2`

`S(4)=36+512-16(16)`

`S(4)=36+512-256`

`S(4)=292`

`S(0)=36+128(0)-16(0)^2`

`S(0)=36`

`\text{Average velocity in first 4 seconds}=(292-36)/(4)`

`\text{Average velocity in first 4 seconds}=(256)/(4)`

`\text{Average velocity in first 4 seconds}=64`

Therefore, the average velocity is first 4 seconds would be 64 feet per second.

(b).
`\text{Average velocity in next 4 seconds}=(S(8)-S(4))/(8-4)`

`S(8)=36+128(8)-16(8)^2`

`S(8)=36+1024-16(64)`

`S(8)=36+1024-1024`

`S(8)=36`

`\text{Average velocity in next 4 seconds}=(36-292)/(4)`

`\text{Average velocity in next 4 seconds}=(-256)/(4)`

`\text{Average velocity in next 4 seconds}=-64`

Therefore, the average velocity is next 4 seconds would be -64 feet per second.