Question

Step 1 of 2: If we ignore air resistance, a falling body will fall 16t^2 feet in t seconds. How far will it fall between t= 3 and t=3.2?

Answer 1

Given data:

* The equation of the distance traveled by the body in terms of time is,

`d=16t^2`

Solution:

The distance traveled by the body in time t = 3 s is,

`\begin{gathered} d(3)=16*(3)^2^{} \\ d(3)=144\text{ ft} \end{gathered}`

The distance traveled by the body in time t = 3.2 s is,

`\begin{gathered} d(3.2)=16*3.2^2 \\ d(3.2)=163.84_{}\text{ ft} \end{gathered}`

Thus, the distance traveled between t = 3 s and t = 3.2 s is,

`\begin{gathered} d(3.2)-d(3)=163.84-144 \\ d(3.2)-d(3)=19.84ft_{} \end{gathered}`

Thus, the body falls 19.84 ft or approx 20 ft in between 3 s and 3.2 s.