A particle moves according to the law `x=r cos((pit)/(2))`. The distance covered by it in the time interval between t = 0 and t = 3 s in nr. What is the value of n?

Answer

To solve the problem, we need to find the distance covered by a particle moving according to the equation \( x = r \cos\left(\frac{\pi t}{2}\right) \) in the time interval from \( t = 0 \) to \( t = 3 \) seconds.

### Step-by-Step Solution:

1. **Identify the displacement function**:
The displacement of the particle is given by:
\[ x(t) = r \cos\left(\frac{\pi t}{2}\right) \]

2. **Find the velocity**:
The velocity \( v(t) \) is the derivative of the displacement with respect to time:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(r \cos\left(\frac{\pi t}{2}\right)\right) \]
Using the chain rule:
\[ v(t) = -r \sin\left(\frac{\pi t}{2}\right) \cdot \frac{\pi}{2} = -\frac{\pi r}{2} \sin\left(\frac{\pi t}{2}\right) \]

3. **Determine the distance covered**:
The distance covered is the integral of the absolute value of velocity over the time interval from \( t = 0 \) to \( t = 3 \):
\[ d = \int_0^3 |v(t)| \, dt = \int_0^3 \left| -\frac{\pi r}{2} \sin\left(\frac{\pi t}{2}\right) \right| dt = \frac{\pi r}{2} \int_0^3 |\sin\left(\frac{\pi t}{2}\right)| dt \]

4. **Analyze the sine function**:
The sine function \( \sin\left(\frac{\pi t}{2}\right) \) oscillates between 0 and 1 in the interval \( [0, 3] \):
- At \( t = 0 \), \( \sin(0) = 0 \)
- At \( t = 1 \), \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- At \( t = 2 \), \( \sin(\pi) = 0 \)
- At \( t = 3 \), \( \sin\left(\frac{3\pi}{2}\right) = -1 \)

Therefore, we can break the integral into two parts:
\[ d = \frac{\pi r}{2} \left( \int_0^2 \sin\left(\frac{\pi t}{2}\right) dt + \int_2^3 -\sin\left(\frac{\pi t}{2}\right) dt \right) \]

5. **Calculate the first integral** \( I_1 \):
\[ I_1 = \int_0^2 \sin\left(\frac{\pi t}{2}\right) dt \]
Let \( u = \frac{\pi t}{2} \) then \( du = \frac{\pi}{2} dt \) or \( dt = \frac{2}{\pi} du \).
Changing the limits:
- When \( t = 0 \), \( u = 0 \)
- When \( t = 2 \), \( u = \pi \)
\[ I_1 = \int_0^\pi \sin(u) \cdot \frac{2}{\pi} du = \frac{2}{\pi} \left[-\cos(u)\right]_0^\pi = \frac{2}{\pi} \left[-(-1 - 1)\right] = \frac{4}{\pi} \]

6. **Calculate the second integral** \( I_2 \):
\[ I_2 = \int_2^3 -\sin\left(\frac{\pi t}{2}\right) dt \]
Using the same substitution:
\[ I_2 = -\int_{\pi}^{\frac{3\pi}{2}} \sin(u) \cdot \frac{2}{\pi} du = -\frac{2}{\pi} \left[-\cos(u)\right]_{\pi}^{\frac{3\pi}{2}} = -\frac{2}{\pi} \left[-0 - 1\right] = \frac{2}{\pi} \]

7. **Combine the integrals**:
\[ d = \frac{\pi r}{2} \left( I_1 + I_2 \right) = \frac{\pi r}{2} \left( \frac{4}{\pi} + \frac{2}{\pi} \right) = \frac{\pi r}{2} \cdot \frac{6}{\pi} = 3r \]

8. **Conclusion**:
The distance covered by the particle in the time interval from \( t = 0 \) to \( t = 3 \) seconds is \( 3r \). Thus, \( n \) is:
\[ n = 3 \]