At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h, and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 P.M.?
mark position A and B, where AB = 150
At a time of t hrs, place the shipA at a point P, between A and B, and place shipB at Q, where Q is vertically above B
so , using Using distance = rate x time
AP = 35t, and BQ = 25t
PBQ is a right-angled triangle
then
PQ^2 = (150 - 35t)^2 + (25t)2
2 PQ d(PQ)/dt = 2(150-35t)(-35) + 2(25t)(25)
at 4:00 pm, t = 4
then PQ^2 = 10^2 + 100^2
PQ = √10100
in the derivative equation, when t = 4 , and PQ=√10100
2√10100 dPQ/dt = 2(10)(-35) + 5000
dPQ/dt = 4300/2√10100
= appr 21.4 km/h
distance is increasing at appr 21.4 km/h
To find the rate at which the distance between ships A and B is changing at 4:00 P.M., we can use the concept of related rates.
Given:
- Ship A is sailing east at 35 km/h.
- Ship B is sailing north at 25 km/h.
- At noon, Ship A is 150 km west of Ship B.
First, let's determine the position of Ship A at 4:00 P.M. from noon:
Time elapsed: 4:00 P.M. - 12:00 P.M. = 4 hours
Distance traveled by Ship A = Speed × Time
= 35 km/h × 4 hours
= 140 km
Since Ship A was initially 150 km west of Ship B, it has now moved 140 km to the east, resulting in a net displacement of 150 km - 140 km = 10 km to the west of Ship B.
Now, we need to find the distance between the ships. This can be done using the Pythagorean theorem:
Distance = √(Side A^2 + Side B^2)
Using the given information, Side A = 10 km (west) and Side B = 150 km (north):
Distance = √(10 km)^2 + (150 km)^2
= √(100 km^2 + 22500 km^2)
= √22600 km^2
≈ 150.83 km
Finally, we need to find the rate at which the distance between the ships is changing at 4:00 P.M. We can do this by differentiating the distance formula with respect to time:
Distance = √(10^2 + 150^2)
Differentiating both sides:
d(Distance)/dt = d(√(10^2 + 150^2))/dt
Now, let's differentiate each term separately:
d(Distance)/dt = (1/2)(10^2 + 150^2)^(-1/2) × (2)(0) + (1/2)(10^2 + 150^2)^(-1/2) × (2)(150) × (d(150)/dt)
Since we are interested in finding the rate at which the distance between the ships is changing, we need to find d(150)/dt. Ship B is moving at a rate of 25 km/h northward, so:
d(150)/dt = 25 km/h
Let's substitute this value back into the equation:
d(Distance)/dt = (1/2)(10^2 + 150^2)^(-1/2) × (2)(0) + (1/2)(10^2 + 150^2)^(-1/2) × (2)(150) × (25 km/h)
Simplifying:
d(Distance)/dt = (1/2)(10^2 + 150^2)^(-1/2) × (2)(150) × (25 km/h)
Now, we can calculate the rate at which the distance between the ships is changing at 4:00 P.M.:
d(Distance)/dt = (1/2)(100 + 22500)^(-1/2) × (2)(150) × (25 km/h)
≈ (1/2)(22600)^(-1/2) × (2)(150) × (25 km/h)
≈ (1/2)(0.0207)^(-1/2) × (2)(150) × (25 km/h)
≈ (1/2)(48.27) × (2)(150) × (25 km/h)
≈ 24.14 × 300 × 25 km/h
≈ 180,750 km/h
Therefore, the rate at which the distance between ships A and B is changing at 4:00 P.M. is approximately 180,750 km/h.
To find the rate at which the distance between the ships is changing, we need to use the concept of relative velocity.
At noon, ship A is 150 km west of ship B, which means that ship A and ship B form a right-angled triangle. We can use this triangle to solve the problem.
Let's assume ship A is located at point A and ship B is located at point B. The distance between them is represented by d, and the time elapsed from noon to 4:00 P.M is t.
To solve the problem, we need to find the rate at which the distance between the ships is changing, which is denoted as dd/dt.
First, let's find the positions of ship A and ship B at 4:00 P.M.
Since ship A is sailing east at 35 km/h, it would have traveled 35 km/h * t (4 hours) = 140 km to the east from its initial position. Thus, its new position, denoted as A', would be 150 km + 140 km = 290 km east of the initial position of ship B.
Similarly, since ship B is sailing north at 25 km/h, it would have traveled 25 km/h * t (4 hours) = 100 km to the north. Thus, its new position, denoted as B', would be 100 km north of the initial position of ship B.
Now that we have the positions of A' and B', we can find the length of the new distance d'.
Using the Pythagorean theorem, we have:
d'^2 = (290 km)^2 + (100 km)^2
Simplifying this equation:
d'^2 = 84100 km^2 + 10000 km^2
d'^2 = 94100 km^2
Taking the square root of both sides:
d' ≈ 306.91 km
Now, let's find the rate at which the distance is changing, dd/dt.
To do this, we differentiate both sides of the equation d'^2 = (290 km)^2 + (100 km)^2 with respect to t.
2d' * dd/dt = 0 + 0
2d' * dd/dt = 0
Simplifying this equation:
2d' * dd/dt = 0
dd/dt = 0 / (2d')
dd/dt = 0
Therefore, at 4:00 P.M., the rate at which the distance between the ships is changing is 0 km/h. This means that the distance between the ships is not changing at that time.