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
The period is
⇒
The amplitude of 60 ⇒
The shift of the midline by 65 units upwards ⇒
The final equation is
and the graph is shown in the third picture below
Answer:
C
Step-by-step explanation:
Corresponding parts of similar triangles are proportionate
11,900,000,000 = 1.19 x 10^10