Ship A is sailing due south at 25 km/hr. At the same time, a second ship B, 50km miles south of A, is sailing due east at 20 km/hr.
(a) at what rate are they approaching or separating at the end of one hour?
(b) At what rate are they approaching or separating at the end of two hours?
(c) When do they cease to approach each other and how far apart are they at that instant
Let B start at (0,0). Then the positions of the boats at time t are
A(t) = (0,50-25t)
B(t) = (20t,0)
and the distance d is thus
d^2 = (20t)^2 + (50-25t)^2
= 1025t^2 - 2500t + 2500
2d dd/dt = 2050t - 2500
At t=1, d ~= 32
64 dd/dt = 2050-2500
dd/dt = -7.03 km/hr
At t=2, d = 40
80 dd/dt = 4100-2500
dd/dt = +20
dd/dt=0 when t=1.22 so d = 31.23
To find out the rate at which the two ships are approaching or separating at different time intervals, we can use the concepts of relative velocity and distance. Let's analyze each part of the question step-by-step:
(a) At the end of one hour:
To determine the rate at which the ships are approaching or separating, we need to find the distance between the two ships after one hour. Since Ship A is sailing south and Ship B is sailing east, we can treat their movements as perpendicular.
Distance Ship A travels in 1 hour = Speed × Time = 25 km/hr × 1 hr = 25 km
Using Pythagoras' theorem, we can find the distance between the two ships after one hour:
Distance^2 = (Distance Ship A)^2 + (Distance Ship B)^2
Distance^2 = (25 km)^2 + (50 km)^2
Calculating:
Distance^2 = 625 + 2500
Distance^2 = 3125
Taking the square root:
Distance = √3125
Distance ≈ 55.9 km
Thus, at the end of one hour, the ships are approximately 55.9 km apart.
(b) At the end of two hours:
To determine the new distance between the ships after two hours, we need to calculate the distance Ship A travels in 2 hours and recalculate the total distance.
Distance Ship A travels in 2 hours = Speed × Time = 25 km/hr × 2 hrs = 50 km
Using the same Pythagorean theorem as before:
Distance^2 = (Distance Ship A)^2 + (Distance Ship B)^2
Distance^2 = (50 km + 25 km)^2 + (50 km)^2
Calculating:
Distance^2 = 75^2 + 2500
Distance^2 = 5625 + 2500
Distance^2 = 8125
Taking the square root:
Distance = √8125
Distance ≈ 90.2 km
Thus, at the end of two hours, the ships are approximately 90.2 km apart.
(c) To determine when the ships cease to approach each other and their distance at that instant:
Since Ship A is moving south at a constant speed and Ship B is moving east at a constant speed, they will eventually reach a point where they no longer approach each other. At this point, the distance between them will remain constant.
Let's say it takes t hours for them to stop approaching each other. During this time, Ship A will have traveled 25t km, and Ship B will have traveled 20t km.
Using the Pythagorean theorem:
Distance^2 = (25t)^2 + (50 - 20t)^2
Distance^2 = 625t^2 + (2500 - 2000t + 400t^2)
Distance^2 = 625t^2 + 2500 - 2000t + 400t^2
Simplifying:
Distance^2 = 1025t^2 - 2000t + 2500
To find when the ships cease to approach each other, we need to solve for t when the rate of change of the distance equals 0.
Differentiating both sides of the equation with respect to t:
d(Distance^2)/dt = 2(1025t^2 - 2000t + 2500) = 0
Simplifying:
2050t - 2000 = 0
Solving for t:
2050t = 2000
t = 2000/2050
t ≈ 0.976 hours (rounded to 3 decimal places)
Therefore, the ships cease to approach each other after approximately 0.976 hours (or about 58.6 minutes). To find the distance at that instant, substitute this value of t into the original equation:
Distance^2 = 625(0.976)^2 + 2500 - 2000(0.976) + 400(0.976)^2
Calculating:
Distance^2 ≈ 4,810.636
Distance ≈ √4,810.636
Distance ≈ 69.4 km
Thus, when the ships cease to approach each other, they are approximately 69.4 km apart.