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

Answer:

(a) 64 feet per second.

(b) -64 feet per second.

Step-by-step 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.

$\text{Average rate change}=\frac{f(b)-f(a)}{b-a}$

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

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

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

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

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

$S(4)=292$

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

$S(0)=36$

$\text{Average velocity in first 4 seconds}=\frac{292-36}{4}$

$\text{Average velocity in first 4 seconds}=\frac{256}{4}$

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

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

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

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

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

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

$S(8)=36$

$\text{Average velocity in next 4 seconds}=\frac{36-292}{4}$

$\text{Average velocity in next 4 seconds}=\frac{-256}{4}$

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

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