Two people start from rest at the same point. One walks east at 3 mi/h and the other walks northeast at 2 mi/h. How fast is the distance between the people changingg after 15 minutes?
one way, using the law of cosines. Draw a triangle. The angle between the two journeys is 45°. At time t, the distances traveled are 3t and 2t. So, the distance x is
x^2 = (3t)^2 + (2t)^2 - 2(3t)(2t)*1/√2
x^2 = 9t^2 + 4t^2 - 6√2 t^2
x^2 = (13-6√2)t^2
At t=15, x = 15√(13-6√2)
2x dx/dt = 2(13-6√2)t
2*15√(13-6√2) dx/dt = 30(13-6√2)
dx/dt = √(13-6√2)
using vectors,
x = (3t,0) - (√2t,√2t) = ((3-√2)t,√2t)
dx/dt = (3√2,-√2)
|dx/dt|^2 = (3-√2)^2 + √2^2 = 9-6√2+2+2 = 13-6√2
|dx|dt| = √(13-6√2)
To find the rate at which the distance between the two people is changing, we can use the concept of relative motion. Relative motion involves finding the difference between the velocities of two objects to determine their relative motion in relation to each other.
Given:
- Person A is moving east at 3 mi/h.
- Person B is moving northeast at 2 mi/h.
To find the rate at which the distance between the two people is changing, we need to determine their relative velocity. We can do this by finding the vector sum of their velocities:
Person A's velocity (vA) = 3 mi/h to the east
Person B's velocity (vB) = 2 mi/h to the northeast
The northeast direction can be decomposed into east and north components. Since the northeast direction forms a 45-degree angle with the east direction, the northeast component is the same as the east component, which is 1 mi/h (found using trigonometry). Therefore, we can express Person B's velocity as:
vB = 1 mi/h to the north + 1 mi/h to the east
Next, we find the relative velocity (vR) between Person A and Person B by subtracting their velocities:
vR = vB - vA
vR = (1 mi/h north + 1 mi/h east) - 3 mi/h east
vR = 1 mi/h north - 2 mi/h east
Now that we have the relative velocity vector, we can determine the rate at which the distance between the two people is changing. Since we are interested in the rate after 15 minutes, we need to convert the time interval to hours:
15 minutes = 15/60 = 1/4 hours
The rate at which the distance is changing is given by the magnitude of the relative velocity multiplied by the time:
Rate = |vR| * (1/4 hours)
Now, let's calculate the magnitude of the relative velocity:
|vR| = sqrt((1 mi/h)^2 + (-2 mi/h)^2)
|vR| = sqrt(1 mi/h + 4 mi^2/h^2)
|vR| = sqrt(5 mi^2/h^2)
Finally, we can find the rate at which the distance is changing by substituting the values into the equation:
Rate = sqrt(5 mi^2/h^2) * (1/4 hours)
After performing the calculation, we can determine the rate at which the distance between the two people is changing after 15 minutes.