A point moves according `s= (2)/(9) "sin" (pi)/(2) t + s_(0)`. The acceleration at the end of first second is

Options

A. `-(pi)/(18)`
B. `-(pi^(2))/(18)`
C. `(pi)/(18)`
D. `(pi^(2))/(18)`

Answer

To find the acceleration of the point at the end of the first second, we start with the given position function:

\[ s = \frac{2}{9} \sin\left(\frac{\pi}{2} t\right) + s_0 \]

**Step 1: Differentiate the position function to find the velocity.**

The velocity \( v \) is the first derivative of the position \( s \) with respect to time \( t \):

\[ v = \frac{ds}{dt} = \frac{d}{dt}\left(\frac{2}{9} \sin\left(\frac{\pi}{2} t\right) + s_0\right) \]

Since \( s_0 \) is a constant, its derivative is zero. Now, we differentiate the sine function:

Using the chain rule, we have:

\[ \frac{d}{dt} \sin\left(\frac{\pi}{2} t\right) = \cos\left(\frac{\pi}{2} t\right) \cdot \frac{d}{dt}\left(\frac{\pi}{2} t\right) = \cos\left(\frac{\pi}{2} t\right) \cdot \frac{\pi}{2} \]

Thus, the velocity becomes:

\[ v = \frac{2}{9} \cdot \frac{\pi}{2} \cos\left(\frac{\pi}{2} t\right) = \frac{\pi}{9} \cos\left(\frac{\pi}{2} t\right) \]

**Step 2: Differentiate the velocity function to find the acceleration.**

The acceleration \( a \) is the derivative of the velocity \( v \) with respect to time \( t \):

\[ a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{\pi}{9} \cos\left(\frac{\pi}{2} t\right)\right) \]

Again using the chain rule, we differentiate the cosine function:

\[ \frac{d}{dt} \cos\left(\frac{\pi}{2} t\right) = -\sin\left(\frac{\pi}{2} t\right) \cdot \frac{d}{dt}\left(\frac{\pi}{2} t\right) = -\sin\left(\frac{\pi}{2} t\right) \cdot \frac{\pi}{2} \]

Thus, the acceleration becomes:

\[ a = \frac{\pi}{9} \cdot \left(-\sin\left(\frac{\pi}{2} t\right) \cdot \frac{\pi}{2}\right) = -\frac{\pi^2}{18} \sin\left(\frac{\pi}{2} t\right) \]

**Step 3: Evaluate the acceleration at \( t = 1 \) second.**

Now, we substitute \( t = 1 \) into the acceleration function:

\[ a(1) = -\frac{\pi^2}{18} \sin\left(\frac{\pi}{2} \cdot 1\right) = -\frac{\pi^2}{18} \sin\left(\frac{\pi}{2}\right) \]

Since \( \sin\left(\frac{\pi}{2}\right) = 1 \):

\[ a(1) = -\frac{\pi^2}{18} \cdot 1 = -\frac{\pi^2}{18} \]

Thus, the acceleration at the end of the first second is:

\[ \boxed{-\frac{\pi^2}{18}} \]

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