Question

Boat A leaves a dock headed due east at 2:00PM traveling at a speed of 9 mi/hr. At the same time, Boat B leaves the same dock traveling due south at a speed of 15 mi/hr. Find an equation that represents the distance d in miles between the boats and any time t in hours.

Answer 1

To find the equation representing the distance between the boats at any time, we can use the concept of relative velocity and the Pythagorean theorem. The equation d = sqrt(306t^2) represents the distance between the boats as a function of time.

Step-by-step explanation:

To find an equation that represents the distance between the boats at any time, we will use the concept of relative velocity. Boat A is moving east with a velocity of 9 mi/hr, and Boat B is moving south with a velocity of 15 mi/hr. The distance between the boats can be represented by the hypotenuse of a right triangle formed by their velocities.

Let's assume the time elapsed is t hours. The distance traveled by Boat A is 9t miles (since it is moving east with a constant velocity). The distance traveled by Boat B is 15t miles (since it is moving south with a constant velocity). Using the Pythagorean theorem, the distance d between the boats can be found using the equation: d^2 = (9t)^2 + (15t)^2.

Simplifying this equation gives us: d^2 = 81t^2 + 225t^2 = 306t^2.

Taking the square root of both sides, we get: d = sqrt(306t^2).

Answer 2

Answer:
`17.5t`

Step-by-step explanation:

Given

Speed of boat A is
`v_a=9\ mi/hr`

Speed of boat B is
`v_b=15\ mi/hr`

Both are moving perpendicular to each other

Distance traveled by Boat A
`x_a=9t`

Distance traveled by Boat B
`x_b=15t`

Distance between them is given by Pythagoras theorem

`\Rightarrow d^2=(x_a)^2+(x_b)^2\\\\\Rightarrow d^2=(9t)^2+(15t)^2\\\\\Rightarrow d=√((9t)^2+(15t)^2)\\\\\Rightarrow d=√(81t^2+225t^2)\\\\\Rightarrow d=√(306t^2)\\\\\Rightarrow d=17.49t\approx 17.5t\ \text{miles}`

Distance between them is
`17.5t`