You bicycle along a straight flat road with a safety light attached to one foot. Your bike moves at a speed of 15 km/hr and your foot moves in a circle of radius 20 cm centered 30 cm above the ground, making one revolution per second.
a. Find parametric equations for x and y which describe the path traced out by the light, where y is distance (in cm) above the ground and x the horizontal distance (in cm) starting position of the center of the circle around which your foot moves. Assuming the light starts 30 cm above the ground, at the front of its rotation. x(t)= , y(t)= . On a separate sheet of paper, sketch the path that your equations describe. b. How fast (in revolutions/sec) would your foot have to be rotating if an observer standing at the side of the road sees the light moving backward? Rotate at revolutions/second.

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.

the answer is two. hope this !

step-by-step explanation:

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