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secondary 4 | A Maths
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Help Naren! Anyone can contribute an answer, even non-tutors.
Help me for question 3(b)
k = 105
(k is the vertical shift of the centre point from the horizontal x-axis)
Since the radius of the wheel is 75 m,
a = 75
(a is the amplitude, or the distance between the highest point and the centre point)
Since it takes 30 minutes to complete one cycle,
Period = 30 minutes
b = Frequency,
b = 2pi / 30
b = pi/15
So, y = 105 + 75 sin (pi/15 t)
When the visitor is 150 m above ground.
105 + 75 sin (pi/15 t) = 150
75 sin (pi/15 t) = 45
sin (pi/15 t) = 3/5
Basic angle = sininv (3/5) = 0.6435011088 rad
t = first quadrant OR second quadrant
t = basic OR pi - basic
t = 0.6435... - 2.498...
So, the person would be 150 m above the ground at times t = 0.64 s amd t = 2.50 s.