Help with trig hw? Please help.
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Help with trig hw? Please help.

[From: ] [author: ] [Date: 11-11-05] [Hit: ]
Make sure you remember that you are in degrees (and not radians).Hope that helps!......

sinθ = opposite/hypotenuse

In this case the "opposite" is the y-coordinate of the rider (compared to the center of the wheel), and the "hypotenuse" is the radius of the wheel (10). θ is the angle from the center of the circle out to where the rider is. So you get:

sinθ = y/10

y = 10*sinθ

When you are dealing with movement around a circle you measure the angles from the right side (the 3 on a clock) at 0 degrees and move counterclockwise. But, we are starting our rider at the bottom of the circle, so we will have to subtract 90 degrees from our angle. Now we have:

y = 10*sin(θ - 90)

This will give you the y-value in relation to the center of the ferris wheel, but we want it in relation to the ground. The center is 11 meters above the ground (radius of 10 meters plus 1 meter off the ground), so we need to ad 11 to our equation. Now we have:

y = 10*sin(θ - 90) + 11

But we don't want an equation in terms of the angle θ, we want it in terms of time. Assuming the rider is moving at a constant speed, and it takes 10 seconds to get half way around the circle, that means the rider can go 180 degrees in 10 seconds. That's a speed of 18 degrees per second. Speed equals distance/time so:

18 = degrees/second = θ / t

θ = 18t

Sub that into our equation and we get:

y = 10*sin(18t-90) + 11

This will give you a height (y, in meters) above the ground at any time (t, in seconds). Make sure you remember that you are in degrees (and not radians).

Hope that helps!
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