The function t=[(sqrt)(d/16)] models the time (t) in seconds that an object has been falling after the object has fallen d feet. When will the time be more than 1 minute?
• d ≥ 16 ft
• d ≥ 225 ft
• d ≥ 3,600 ft
• d ≥ 57,600 ft
t=√(d/16)
so you want √(d/16) > 1
square both sides
d/16 > 1
d > 16
D. d>57600
Why did the object want to fall for more than 1 minute? Because it wanted to get a longer break from gravity's constant nagging!
To find when the time is more than 1 minute, we can plug in the values and see when the inequality is true:
1. For d ≥ 16 ft:
Let's plug in d = 16 ft into the equation:
t = (√(16/16))
t = (√1)
t = 1 second
Uh-oh, it's less than 1 minute. So, the time is not more than 1 minute when d ≥ 16 ft.
2. For d ≥ 225 ft:
Let's plug in d = 225 ft into the equation:
t = (√(225/16))
t = (√14.06)
t ≈ 3.75 seconds
Still not over 1 minute. So, the time is not more than 1 minute when d ≥ 225 ft.
3. For d ≥ 3,600 ft:
Let's plug in d = 3,600 ft into the equation:
t = (√(3600/16))
t = (√225)
t = 15 seconds
Well, it's still not more than 1 minute. So, the time is not more than 1 minute when d ≥ 3,600 ft.
4. For d ≥ 57,600 ft:
Let's plug in d = 57,600 ft into the equation:
t = (√(57600/16))
t = (√3600)
t = 60 seconds
Aha! Finally, when d is greater than or equal to 57,600 ft, the time will be more than 1 minute.
To find when the time will be more than 1 minute, we need to solve the inequality t > 1.
Let's solve this inequality for each given condition:
1. For d ≥ 16 ft:
Substituting the given function t = (√d/16), we have:
(√d/16) > 1
Squaring both sides of the inequality:
d/16 > 1
Multiplying both sides by 16:
d > 16
So, for d ≥ 16 ft, the time will be more than 1 minute.
2. For d ≥ 225 ft:
Substituting the given function t = (√d/16), we have:
(√d/16) > 1
Squaring both sides of the inequality:
d/16 > 1
Multiplying both sides by 16:
d > 16
Since 16 is less than 225, this condition does not change the result from the previous condition. Therefore, for d ≥ 225 ft, the time will also be more than 1 minute.
3. For d ≥ 3,600 ft:
Substituting the given function t = (√d/16), we have:
(√d/16) > 1
Squaring both sides of the inequality:
d/16 > 1
Multiplying both sides by 16:
d > 16
Again, 16 is less than 3,600, so this condition does not change the result. Therefore, for d ≥ 3,600 ft, the time will also be more than 1 minute.
4. For d ≥ 57,600 ft:
Substituting the given function t = (√d/16), we have:
(√d/16) > 1
Squaring both sides of the inequality:
d/16 > 1
Multiplying both sides by 16:
d > 16
Once again, 16 is less than 57,600, so this condition does not change the result. Therefore, for d ≥ 57,600 ft, the time will also be more than 1 minute.
In conclusion, for all four given conditions (d ≥ 16 ft, d ≥ 225 ft, d ≥ 3,600 ft, and d ≥ 57,600 ft), the time will be more than 1 minute.
To determine when the time will be more than 1 minute, we need to solve the inequality:
t > 1 minute
First, let's convert 1 minute to seconds. Since there are 60 seconds in 1 minute, 1 minute is equal to 60 seconds.
t > 60 seconds
Now, substitute the given equation for t:
[(sqrt)(d/16)] > 60
To solve this inequality, we need to isolate the variable d.
Step 1: Square both sides of the inequality to remove the square root:
[(sqrt)(d/16)]² > 60²
Simplifying this gives us:
d/16 > 3600
Step 2: Multiply both sides of the inequality by 16 to eliminate the fraction:
d > 3600 * 16
Simplifying this gives us:
d > 57600
Therefore, the correct option is:
d ≥ 57,600 ft