You bicycle along a straight flat road with a safety light attached to one foot. Your bike moves at a speed of 10 km/hr and your foot moves in a circle of radius 24 cm centered 34 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 cm above the ground, at the front of its rotation.
x(t)=
y(t)=

(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) = 2.78t + 0.24cos 2πt

& Y(t) = 0.24 - 0.24sin2πt

B) 1000(n-0.5) ≤ r ≤ 1000n

Foe values of n at 1, and 2, the speed is between 500 & 1000 and between 1500 & 2000 respectively. For higher values of n, we can calculate the range of r from the formula.

Step-by-step explanation:

To find the parametric equation, we need to find the vector for horizontal and vertical movements. Let's call it (x(t), y(t))

Since there is no vertical movement of bicycle, the initial component is;

(0, 0.24) (since it's centred 34cm above ground)

Now let's convert 10km/h to m/s;

Hence, 10km/h = 2.78 m/s

So vector of this in horizontal direction is; (2.78t,0)

Now, the circular motion component for vector with radius 0.24m is; (0.24cos 2πt, - 0.24Sin 2πt).

Now, when we add the 3 components to get;

X(t) = 0 + 2.78t + 0.24cos 2πt

= 2.78t + 0.24cos 2πt

And Y(t) = 0.24 + 0 - 0.24sin2πt

= 0.24 - 0.24sin2πt

For motion to appear backwards, we should only see half a revolution in his eye period. Therefore, for this, n ≥ 1, eye period is (n-0.5)P

Where P is the light period.

From research, human eye can see up to 1000 frames per second; therefore, converting it to seconds per frame, we have ; 1/1000 = 0.001

Hence,

(n-0.5)P ≤ 0.001 ≤ nP

Multiplying each term by 1000/P to get;

1000(n-0.5) ≤ 1/P ≤ 1000n

1/P is called rate of revolution (r)

So, 1000(n-0.5) ≤ r ≤ 1000n

So that's the formula when n ≥ 1.

When n is 1, r is between 500 and 100. When n is 2, r is between 1500 and 2000. And so it goes for values of n greater than 1.

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step-by-step explanation:

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: