Answer:
a)
x(t) = ( 4.167 t + 0.2 cos (2πt))
y(t) = ( 0.3 - 0.2 sin (2πt))
b) the foot have to be 3.32 rev/sec faster
Step-by-step explanation:
Given that:
the speed of the bike = 15 km/hr = 15 × 1000/3600 (m/sec) = 4.167 m/sec
radius of the circle when the foot moves = 20 cm = 0.2 m
radius of the circle above the ground = 30 cm = 0.3 m
Let assume that:
x(t) should represent the vector along the horizontal moment
y(t) should be the vector along the vertical moment
The initial component will be ( 0, 0.3)
We know that the radius of the circle is given as 0.2 m, So the vector of the circle can be written as (0.2 cos t , 0.2 sin t )
Also, the foot makes one revolution in a second, definitely the frequency of the revolution = 1 and the vector for the circle is ( 0.2 cos (2πt), -0.2 sin (2πt)), due to the fact that the foot moves clockwise.
Thus, adding all the component together ; we have:
(x(t), y(t)) = (0,0.3)+(4.167 t , 0)+(0.2 cos (2πt), -0.2 sin (2πt))
(x(t), y(t)) = (4.167 t + 0.2 cos (2πt), 0.3 - 0.2 sin (2πt))
Hence; the parametric equations are:
x(t) = ( 4.167 t + 0.2 cos (2πt))
y(t) = ( 0.3 - 0.2 sin (2πt))
b)
The linear speed of rotation is :
15km/hr = 15 × 100, 000/3600 (cm/sec)
= 416.7 cm/sec
The rotational frequency is :
= 416.7/2πr
= 416.7/2(3.14 × 20)
= 3.32 rev/sec
Hence, the foot have to be 3.32 rev/sec faster in rotating if an observer standing at the side of the road sees the light moving backward.
