Two poles are 24 metres and 30 metres high and 20 metres apart. A slack wire joins the tops of the poles and is 32 metres long. A cable car is moving along the wire at 5 metres per second away from the shorter pole. When the car is 12 metres horizontally from the shorter pole, it is 15 metres high and the length of the wire to the shorter pole is 15 metres. At this instant, find: a) how fast the car is moving horizontally. b) how fast the car is moving vertically.
No ideas? OK. I'll get you started. As usual, with a diagram.
Let's set some labels:
S = top of short pole
T = top of tall pole
C = cable car
P = point on ground directly below the car
Draw a horizontal line from C which intersects the short and tall poles at Q and R, respectively.
B = bottom of short pole
D = bottom of tall pole.
x = BP = QC (hor. distance of car from short pole)
y = QS (distance of car below short pole)
RT = y+6 (distance of car below tall pole)
u = CS (length of cable from short pole)
CT = 32-u (cable length to tall pole)
That means we have at the moment in question,
x=12
y=9
z=15
u=15
v=17
Wow - notice how all the distance come out to be integers? Well crafted problem!
We want to find dx/dt and dy/dt.
x^2+y^2 = u^2
(20-x)^2 + (y+6)^2 = (32-u)^2
See what you can do with that.
To find the rate at which the car is moving horizontally (a) and vertically (b), we can use the concepts of similar triangles and the Pythagorean theorem.
Let's start by drawing a diagram to visualize the situation:
```
B (30m)
/ |
/ |
/ |
/ |
/ |
/ |
/ |
/ |
/ |
/ |
A (24m) /___________C (32m)
20m
```
In the diagram above, A represents the shorter pole, B represents the taller pole, and C represents the position of the cable car. The letters in parentheses indicate the heights of the respective points.
Now, let's consider the triangles in this diagram. Triangle ACB and triangle CDX (where D is the intersection of AC with the wire and X is the position of the cable car at that instant) are similar triangles, since they share the same angles.
Using the similar triangles, we can set up the following proportions:
AB / AC = BX / CX (1) -- This relates the horizontal distances.
AB / AC = BC / CD (2) -- This relates the vertical heights.
From equation (1), we can rewrite it as:
20 / 32 = BX / 12
Simplifying, we have:
BX = (20 / 32) * 12 = 7.5 meters
So, the car is moving horizontally at a speed of 7.5 meters per second.
Now, let's focus on equation (2):
24 / 32 = 15 / CD
Simplifying, we have:
CD = (32 / 24) * 15 = 20 meters
So, the length of the wire to the shorter pole is 20 meters.
Therefore, the car is moving vertically at a speed of 20 meters per second.
To summarize:
a) The car is moving horizontally at a speed of 7.5 meters per second.
b) The car is moving vertically at a speed of 20 meters per second.