Answer:
a. k = 1.62
b. 40.5 m
c. 11.11 s
Explanation:
Part a.
If d is the distance traveled and t is the seconds the object has been falling, we can represent the direct variation with the following equation
d = kt²
Where k is the constant of variation.
If the object has been falling for 10 seconds and travels 162 m, we can replace t = 10 and d = 162 and solve for k as follows
[tex]\begin{gathered} 162=k(10)^2 \\ 162=k(100) \\ \\ \frac{162}{100}=k \\ \\ 1.62=k \end{gathered}[/tex]Therefore, the constant of variation is 1.62.
Part b.
Then, the equation that related the distance and the square of the time is
d = 1.62t²
So, we can calculate the distance an object falls in 5 seconds replacing t = 5
d = 1.62(5)²
d = 1.62(25)
d = 40.5 m
So, the distance is 40.5 m
Part c.
To calculate the time required for an object to fall 200 m, we need to replace d = 200 and solve for t, so
[tex]\begin{gathered} 200=1.62t^2 \\ \\ \frac{200}{1.62}=t^2 \\ \\ 123.46=t^2 \\ \sqrt{123.46}=t \\ 11.11\text{ s = t} \end{gathered}[/tex]Therefore, the time required is 11.11 seconds