Question

20 POINTS, PLEASE HELP ASAP!!!!

Answer 1

Question b)

17 minutes ⇒ 1 revolution = 2π
1 minute ⇒ 2π/17

Angle of rotation is 2π/17 per minute

Question c)

The period of the rotation is 2π/17 

The wheel starts rotating from the height of 5 meters from the ground and the maximum height is 125 meter from the ground.
The amplitude is 125 - 5 = 120 ÷ 2 = 60

The midline of the rotation is at 125 - 5= 120 ÷2 = 60 + 5 = 65 (this is the location of the centre of the wheel). This value is a shift of 65 from the zero midlines (which in this case would be the ground)

Question C

The movement of the wheel can be described as starting from 0° then reaching its peak at 180° and come back to its original position as it stops and it makes a complete circular turn 360°.

If we sketch this on a graph, we will obtain a curve of -cos(x)

We need to apply the period, the amplitude and the shift produced by this rotation into the equation $y = -cos(x)$

The period is $\frac{2 \pi }{7}$ ⇒ $f(t)=-cos( \frac{2 \pi }{7}t)$
The amplitude of 60 ⇒ $f(t)=-60cos( \frac{2 \pi }{7}t)$
The shift of the midline by 65 units upwards ⇒ $f(t) = -60cos( \frac{2 \pi }{7}t)+65$

The final equation is
$f(t) = -60cos( \frac{2 \pi }{7}t)+65$

and the graph is shown in the third picture below