trigonometry problem

**Jasleen2008** · December 5th 2008, 12:23 AM

A ferris wheel has a radius of 160 feet rotates at a constant 1.57 revolutions per minute. The lowest point in the wheel is 24 feet off the ground. From the point where you start the ride, it takes you 21 seconds to reach the highest point on the wheel.

a) How far off the ground will you be when the ride starts?

b) How far off the ground will you be after 5.6 minutes?

**nzmathman** · December 5th 2008, 11:03 AM

This situation can be modeled by a trigonometric equation.

We will use cos.

$H = A\cos{(\omega t - D)} + C$

A is the amplitude; which is 160ft.

$\omega t$ is the period.

$\omega = \frac{\theta}{t}$

$\omega = \frac{1.57 \times 2\pi radians}{60} = 0.1644$

C is the vertical translation. The centre is now at $24ft + 160ft = 184ft$ high.

So $C = 184$

Therefore

$H = 160\cos{(0.1644t-D)} + 184$

To solve D use the info given (at 21secs you are at max height 344ft)

$344 = 160\cos{(0.1644\times21-D)} +184$

$344 = 160\cos{(0.1644\times21-D)} +184$

$D= 3.4524$

The full equation is therefore:

$H = 160\cos{(0.1644t-3.4524)} + 184$

Q (a)

Substitute $t = 0$ into equation for height.

Q (b)

Substitute $t = 5.6 \times 60 = 336$ into equation for height.

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