Ronan is roamin up and down the corridor on a unicycle. His Displacement from the middle of the corridor is given by x = 12cos(2t), where x is in metres and t is in seconds. After $$ \frac{49\pi}{6} $$seconds his unicycle breaks down. After $$ \frac{\pi}{2} $$ seconds, Ronan is moving again. He immediately continues his motion pattern from where it had been paused. What is Ronan's displacement and velocity $$ \frac{41\pi}{4} $$seconds after initially commencing movement

Given that Ronan's displacement from the middle of the corridor is given by $ x(t) = 12 \cos(2t) $, where $ x $ is in meters and $ t $ is in seconds, we need to find his displacement and velocity $ \frac{41\pi}{4} $ seconds after initially commencing movement.

### 1. Understand the problem

The displacement function is:

\[
x(t) = 12 \cos(2t)
\]

The velocity is the derivative of the displacement with respect to time $ t $:

\[
v(t) = \frac{dx(t)}{dt} = -24 \sin(2t)
\]

### 2. Determine the total time elapsed after the pause

Ronan’s unicycle breaks down after $ \frac{49\pi}{6} $ seconds and resumes after $ \frac{\pi}{2} $ seconds. Therefore, the total time elapsed when he resumes motion is:

\[
t_{\text{resume}} = \frac{49\pi}{6} + \frac{\pi}{2} = \frac{49\pi}{6} + \frac{3\pi}{6} = \frac{52\pi}{6} = \frac{26\pi}{3} \text{ seconds}
\]

Given that we want to find the displacement and velocity at $ \frac{41\pi}{4} $ seconds after initially commencing movement, we need to calculate the additional time after the unicycle resumes:

\[
\text{Total time} = \frac{41\pi}{4}
\]

Subtract the time at which he resumed:

\[
t_{\text{after resume}} = \frac{41\pi}{4} - \frac{26\pi}{3}
\]

First, let's rewrite $ \frac{26\pi}{3} $ with a common denominator:

\[
\frac{26\pi}{3} = \frac{104\pi}{12}
\quad \text{and} \quad
\frac{41\pi}{4} = \frac{123\pi}{12}
\]

Thus:

\[
t_{\text{after resume}} = \frac{123\pi}{12} - \frac{104\pi}{12} = \frac{19\pi}{12} \text{ seconds}
\]

### 3. Find displacement and velocity

Now, Ronan continues his motion from where it paused, so his displacement $ x $ at $ t_{\text{after resume}} $ is:

\[
x\left(\frac{19\pi}{12}\right) = 12 \cos\left(2 \times \frac{19\pi}{12}\right) = 12 \cos\left(\frac{19\pi}{6}\right)
\]

Since $ \cos(\theta) = \cos(\theta + 2\pi k) $ for any integer $ k $, we can simplify $ \frac{19\pi}{6} $:

\[
\frac{19\pi}{6} = \frac{3\pi}{6} + 3\pi = \frac{\pi}{2} + 3\pi
\]

\[
x\left(\frac{19\pi}{12}\right) = 12 \cos\left(\frac{\pi}{2} + 3\pi\right) = 12 \cos\left(\frac{\pi}{2} \right) = 12 \times 0 = 0
\]

Thus, **Ronan’s displacement is 0 meters**.

Now, find the velocity at this time:

\[
v\left(\frac{19\pi}{12}\right) = -24 \sin\left(\frac{19\pi}{6}\right) = -24 \sin\left(\frac{3\pi}{6} + 3\pi\right) = -24 \sin\left(\frac{\pi}{2} + 3\pi\right)
\]

Since $ \sin(\theta + \pi) = -\sin(\theta) $:

\[
v\left(\frac{19\pi}{12}\right) = -24 \times (-1) = 24 \text{ m/s}
\]

Thus, **Ronan’s velocity is 24 m/s**.

Would you like more details or have any questions?

Here are 5 related questions for practice:
1. What is Ronan's displacement $ \frac{\pi}{4} $ seconds after he starts moving?
2. How would the displacement change if the equation was $ x(t) = 12 \sin(2t) $?
3. What is the maximum velocity Ronan can achieve?
4. After how much time will Ronan first return to his starting point?
5. What is the velocity at $ t = \frac{25\pi}{4} $ seconds?

**Tip:** When analyzing sinusoidal functions, remember to use the properties of sine and cosine to simplify expressions.

Explore Ronan's displacement and velocity problem where his motion is described by x = 12cos(2t). Learn how to calculate his displacement and velocity at 41π/4 seconds after initially commencing movement. Includes detailed step-by-step solutions and related practice questions.

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