"At 9am ship A is 50 km east of ship B. Ship A is sailing north at 40km/h and ship B is sailing south at 30km/h. How fast is the distance between them changing at noon?"
I can never get these type of questions where two ships are heading in different directions and one is always ahead...please help!
For most of these, the key is to have a good diagram.
In this problem I would place B at the origin and A on the 50 axis somewhere.
label AB as 50
draw a vertical up from A to C to show the path of A
Draw a vertical down from B to D to show the path of B
Join DC, this is our distance between them.
Now we need a righ-angled triangle, so extend CA to E so that AE = BD. Join DE
Let t hours be any time after 9:00 am
then BD = 30t and AC = 40t, making CE = 70t
We know DE = 50
DC^2 = CE^2 + DE^2
= 4900t^2 + 2500
differentiate with respect to t
2 DC (dDC/dt) = 9800t
dDC/dt = 9800t/2DC
when t- 3 (noon)
DC^2 = 4900(9) + 2500
DC = 215.87
so dDC/dt = 9800(3)/(2(215.87)) = 68.1 km/h
This same question was answered Wednesday for your 10:58pm post.
Check back a few pages to see it.
Thank you both! However, the problem is I can't seem to visualize the diagram correctly. I can't see where the right triangle is coming from. But no worries, I will look at more example problems and see if I can figure these questions out.
To solve this type of question, you need to understand the concept of relative velocity. The relative velocity is the velocity of one object with respect to another.
In this problem, let's consider ship A as the reference point. Ship A is sailing north at a speed of 40 km/h. Ship B is sailing south at a speed of 30 km/h.
Now, at a given time, let's say at 9 am, ship A is 50 km to the east of ship B. To find the rate at which the distance between the two ships is changing, we need to find the derivative of the distance between them with respect to time.
First, we can visualize their initial positions on a coordinate grid with ship A at (0, 50) and ship B at (0, 0).
Next, we can determine the position of the ships at any given time using their velocities and the time elapsed. For instance, at noon (3 hours later), ship A will have moved north by 40 km/h * 3 hours = 120 km from its initial position, and ship B will have moved south by 30 km/h * 3 hours = 90 km from its initial position.
Now, draw a right triangle to represent the positions of the ships, with the line connecting ship A and ship B representing the distance between them. The vertical side of the triangle represents the north-south position, which is changing at a rate of (40 km/h - 30 km/h) = 10 km/h. The horizontal side of the triangle represents the east-west position, which remains unchanged.
Using the Pythagorean theorem, we can calculate the distance between the ships at noon:
Distance^2 = (east-west position)^2 + (north-south position)^2
Distance^2 = 50^2 + (120 - 90)^2 = 2500 + 900 = 3400
Distance = sqrt(3400) = 58.31 km
Now, to find the rate of change of the distance between them at noon, we can differentiate the distance equation with respect to time t: d(distance)/dt.
d(distance)/dt = (2 * distance) * (d(distance)/dt)
d(distance)/dt = 2 * sqrt(3400) * (d(distance)/dt)
We need to find d(distance)/dt, which is the rate at which the distance between the two ships is changing at noon. To do this, we multiply the rate of change of the north-south position (10 km/h) by 2, and substitute the distance value:
d(distance)/dt = 2 * sqrt(3400) * (10 km/h)
d(distance)/dt = 2 * 58.31 km * 10 km/h
d(distance)/dt = 1166.2 km/h
Therefore, the distance between the two ships is changing at a rate of 1166.2 km/h at noon.