Parametric question

Question

Two cars are racing at a constant speed around a circular racetrack. Car A requires 15 seconds to travel once around the racetrack, and car B requires 25 seconds to travel once around the racetrack. If car A passes car B, how many seconds will elapse before car A once again passes car B ? This problem can be solved in various techniques, you must use a parametric equation method.

Step 1of 3 Find a set of parametric equations that model the movement described above. Your equations must simulate the movement around a circular track and that each car completes one revolution in the amount of time given. (hint: these equations will likely contain trigonometric functions)

step 2 of 3 Find at what parameter value ( – time) do the two cars meet again. (hint you may need to set your x-coordinates of each parametric equation equal to each other and check y-coordinates, you may need your product-to-sum identities from trigonometry )

step 3 of 3 Now solve the problem using a different technique (ie non-parametric ) Explain your logic and your conclusion.

Answer 1

Step 1: Parametric Equations

Given:

Car A takes 15 seconds to complete one lap.

Car B takes 25 seconds to complete one lap.

The parametric equations for each car are:

1. Car A:

\[

x_A(t) = \cos\left(\frac{2\pi}{15} t\right), \quad y_A(t) = \sin\left(\frac{2\pi}{15} t\right)

\]

2. Car B:

\[

x_B(t) = \cos\left(\frac{2\pi}{25} t\right), \quad y_B(t) = \sin\left(\frac{2\pi}{25} t\right)

\]

Step 2: Finding the Time When the Cars Meet Again

To find the time when Car A passes Car B again, you need to solve the equations for when the positions match. Specifically, we need \( x_A(t) = x_B(t) \) and \( y_A(t) = y_B(t) \).

Setting \( x_A(t) = x_B(t) \):

\[

\cos\left(\frac{2\pi}{15} t\right) = \cos\left(\frac{2\pi}{25} t\right)

\]

And setting \( y_A(t) = y_B(t) \):

\[

\sin\left(\frac{2\pi}{15} t\right) = \sin\left(\frac{2\pi}{25} t\right)

\]

Using the periodic properties of the cosine and sine functions, the two cars meet again when:

\[

\frac{2\pi}{15} t = \frac{2\pi}{25} t + 2k\pi

\]

Solving for \( t \):

\[

\frac{2\pi}{15} t - \frac{2\pi}{25} t = 2k\pi

\]

Factoring out \( 2\pi \):

\[

\left(\frac{1}{15} - \frac{1}{25}\right) t = k

\]

Finding a common denominator:

\[

\left(\frac{5}{75} - \frac{3}{75}\right) t = k

\]

\[

\frac{2}{75} t = k

\]

Solving for \( t \):

\[

t = \frac{75k}{2}

\]

The smallest positive solution is when \( k = 1 \):

\[

t = \frac{75}{2} \text{ seconds}

\]

Step 3: Solving the Problem Using a Different Technique

Another method to find when Car A passes Car B again is by using the concept of relative speed.

The relative speed between Car A and Car B is the difference in their angular velocities. Car A's angular velocity is \( \frac{2\pi}{15} \) radians per second, and Car B's angular velocity is \( \frac{2\pi}{25} \) radians per second.

The relative angular velocity is:

\[

\omega_{relative} = \left| \frac{2\pi}{15} - \frac{2\pi}{25} \right| = 2\pi \left| \frac{1}{15} - \frac{1}{25} \right| = 2\pi \left| \frac{5 - 3}{75} \right| = \frac{2\pi \cdot 2}{75} = \frac{4\pi}{75}

\]

The time it takes for Car A to pass Car B again is the time for one full lap of relative motion, which is when the angle covered by the relative motion is \( 2\pi \):

\[

\text{Time} = \frac{2\pi}{\omega_{relative}} = \frac{2\pi}{\frac{4\pi}{75}} = \frac{75}{2}

\]

Thus, using both parametric and relative speed methods, you find that Car A will pass Car B again after \( \frac{75}{2} \) seconds.

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