Boat A leaves a dock headed due east at 1:00 PM traveling at a speed of 16 mi/hr. At 2:00 PM, Boat B leaves the same dock traveling due north at a speed of 18 mi/hr. Find an equation that represents the distance d in miles between the boats and any time t in hours for t≥1 , using that t=0 corresponds to the time that Boat A leaves the dock.
To find the equation that represents the distance between the boats, we need to consider the motion of each boat separately and then calculate the distance between their positions.
Let's first consider the motion of Boat A. It is traveling due east at a speed of 16 mi/hr. Since t=0 corresponds to the time Boat A leaves the dock, we can say that the position of Boat A at any time t≥0 can be represented by the equation:
x_A = 16t
Here, x_A represents the distance traveled by Boat A in miles.
Now let's consider the motion of Boat B. It is traveling due north at a speed of 18 mi/hr. However, Boat B leaves the dock at t=1, so we need to account for the one-hour time difference. The position of Boat B at any time t≥1 can be represented by the equation:
y_B = 18(t-1)
Here, y_B represents the distance traveled by Boat B in miles.
To find the distance d between the boats at any time t, we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides:
d^2 = x_A^2 + y_B^2
Substituting the equations for x_A and y_B, we have:
d^2 = (16t)^2 + [18(t-1)]^2
Simplifying this equation further, we have:
d^2 = 256t^2 + 324(t-1)^2
Hence, the equation that represents the distance d in miles between the boats and any time t in hours for t≥1 is:
d^2 = 256t^2 + 324(t-1)^2
To find an equation that represents the distance between the boats at any given time, we can break down the problem into smaller steps.
Step 1: Determine the position equations for each boat.
We know that Boat A is moving due east at a speed of 16 mi/hr, and it leaves the dock at 1:00 PM (which corresponds to t=0). The position equation for Boat A can be expressed as:
Position A(t) = 16t
Boat B, on the other hand, is moving due north at a speed of 18 mi/hr and left the dock at 2:00 PM. Since boat B starts one hour later, we need to account for the time difference. Therefore, the position equation for Boat B becomes:
Position B(t) = 18(t-1)
Step 2: Find the distance between the boats.
The distance between the boats can be calculated using the distance formula, which is the square root of the sum of the squares of the horizontal and vertical distances. In this case, the horizontal distance is the difference in eastward positions between the boats, and the vertical distance is the difference in northward positions.
So, the equation for the distance (d) between the boats at any given time (t) can be represented as:
d(t) = sqrt((Position A(t) - Position B(t))^2 + (0 - Position B(t))^2)
Substituting the position equations we derived earlier, we get:
d(t) = sqrt((16t - 18(t-1))^2 + (-18(t-1))^2)
Simplifying this equation will give the final equation that represents the distance between the boats.
We have a right triangle whose hypotenuse is the distance (d) between the boats:
d^2 = 16t^2 + 18(t-1)^2