Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 30 feet long, but can stretch up to 90 feet. They both start from the same location. Allyson moves 20 ft/sec and Adrian moves 12 ft/sec in the directions indicated. (If a coordinate system is used, assume that the girls' starting position is located at (x, y) = (0, 0) and that Allyson and Adrian move in the positive y and negative x directions, respectively. Let one unit equal one foot.)

(a) Where are the two girls located after 2 seconds?
Allyson (x,y)
Adrian (x,y)
(c) Determine when the bungee cord first becomes tight; i.e., there is no slack in the line this = 1.3 sec
Where are the girls located when this occurs? (Round your answers to one decimal place as needed.)
Allyson (x,y)
Adrian (x,y)
can anyone help me

well it says that

Allyson is at (0,20t)
Adrian is at (-12t,0)
is wrong

how does starting at (0,0) and moving north at 20 ft/s not put Allyson at (0,20t) after t seconds? Something's wrong here.

At t=2, Allyson must be at (0,40)

To solve this problem, we need to track the positions of Allyson and Adrian over time. Let's break down the steps:

(a) To find the positions of the girls after 2 seconds, we can use their speeds and the formula:

Position = Initial Position + (Speed * Time)

For Allyson:
Initial Position (x,y) = (0,0) since they start from the same location.
Speed = 20 ft/sec
Time = 2 sec

Using the formula, we can calculate:
Allyson (x,y) = (0 + (20 * 2), 0 + (20 * 2))
= (40, 40)

For Adrian:
Initial Position (x,y) = (0,0) since they start from the same location.
Speed = 12 ft/sec
Time = 2 sec

Using the formula, we can calculate:
Adrian (x,y) = (0 - (12 * 2), 0 + (12 * 2))
= (-24, 24)

Therefore, after 2 seconds:
Allyson is located at (40, 40) and Adrian is located at (-24, 24).

(c) To determine when the bungee cord first becomes tight, we need to find the time at which the distance between the two girls reaches the maximum length of the bungee cord, which is 90 ft.

The distance between them can be calculated using the distance formula:

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

where (x1, y1) and (x2, y2) are the positions of Allyson and Adrian respectively.

We need to solve for the time when the distance becomes 90 ft. So, let's set up the equation:

Distance = 90 ft
sqrt((x2 - x1)^2 + (y2 - y1)^2) = 90

Substituting the positions of Allyson and Adrian at any given time:

sqrt((40 - (-24))^2 + (40 - 24)^2) = 90

Simplifying:
sqrt(64^2 + 16^2) = 90
sqrt(4096 + 256) = 90
sqrt(4352) = 90
(sqrt(2^8 * 17)) = 90

Taking the square root of both sides:
(sqrt(2^8 * 17)) = sqrt(90)^2
16sqrt(17) = 90

Simplifying:
sqrt(17) = 90/16
sqrt(17) ≈ 5.63

Now we can find the time when the distance is 5.63 by using the distance formula:

Time = Distance / Speed

Distance = 5.63 ft
Speed = Relative speed between Allyson (20 ft/sec) and Adrian (12 ft/sec)

Time = 5.63 / (20 + 12)
Time ≈ 0.20825 seconds (rounded to 1 decimal place)

Therefore, the bungee cord first becomes tight after approximately 0.2 seconds.

To determine the positions of the girls at this time, we use the speed and time formulas:

For Allyson:
Initial Position (x,y) = (0,0)
Speed = 20 ft/sec
Time = 0.2 sec

Allyson (x,y) = (0 + (20 * 0.2), 0 + (20 * 0.2))
= (4, 4)

For Adrian:
Initial Position (x,y) = (0,0)
Speed = 12 ft/sec
Time = 0.2 sec

Adrian (x,y) = (0 - (12 * 0.2), 0 + (12 * 0.2))
= (-2.4, 2.4)

Therefore, when the bungee cord first becomes tight (at approximately 0.2 seconds), Allyson is located at (4, 4) and Adrian is located at (-2.4, 2.4) (rounded to 1 decimal place).

As I told you 8 hours ago,

cannot access files on your computer. But, clearly, after t seconds,
Allyson is at (0,20t)
Adrian is at (-12t,0)
so plug in t=2 or 1.3