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 26 cm centered 36 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.
(b) How fast (in revolutions/sec)

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- schrodinger

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- anonymous

your foot have to be rotating if an observer standing at the side of the road sees the light moving backward?
Answer in revolution per second. Thank You!

- BAdhi

you can start by writing \((x,y)\) with \(r\) and \(\theta\) where r is the radius and \(\theta\) is the angle

- anonymous

revolution per second

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## More answers

- anonymous

ok wait, so i got this for x(t) = 26cos(2pi(t)) + 277.8t
and y(t) = -26sin(2pi(t)) + 32 (i'm not sure if its supposed to be positive or negative)
And i have no idea how to do part b. Thx for the help!

- BAdhi

I think constant of the y(t) has to 36
if we take the horizontal velocity,
\[V_x(t) = -26\sin(\omega t) +277.8\]
what they are asking is the value of the \(\omega\) for \(V_x < 0\)

- anonymous

can you help with b plz?

- BAdhi

I think i've made a mistake with the \(V_x\),
It shoud be,
\[V_x(t) = -26\omega \sin(\omega t) + 277.8\]
Now to see the light moving backward, \(V_x \leq 0\) should be fulfilled
\[ -26\omega\sin(\omega t) + 277.8\leq 0 \implies \omega \geq \frac{277.8}{26\sin(\omega t)} \]
To get the minimum requirement, we consider when \(\sin(\omega t)=1\)
So \(\omega \geq 277.8/26\)

- anonymous

its says its not correct tho :/ Thanks for helping me btw!

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