Question

Please help me to answer and explain how you get the answer to number 11. Thank you.

Answer 1

a. k = 1.62

b. 40.5 m

c. 11.11 s

Step-by-step 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

`\begin{gathered} 162=k(10)^2 \\ 162=k(100) \\ \\ (162)/(100)=k \\ \\ 1.62=k \end{gathered}`

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

`\begin{gathered} 200=1.62t^2 \\ \\ (200)/(1.62)=t^2 \\ \\ 123.46=t^2 \\ √(123.46)=t \\ 11.11\text{ s = t} \end{gathered}`

Therefore, the time required is 11.11 seconds