Answer:
Step-by-step explanation:
To find the acceleration of the particle at \(t = 2\), you need to take the derivative of the velocity function \(v(t)\) with respect to time \(t\). The acceleration is the rate of change of velocity.
Given the velocity function \(v(t) = \sin(\pi t)\), let's find the acceleration \(a(t)\):
\[ v(t) = \sin(\pi t) \]
Now, take the derivative of \(v(t)\) with respect to \(t\) to find \(a(t)\):
\[ a(t) = v'(t) = \frac{d}{dt}(\sin(\pi t)) \]
The derivative of \(\sin(\pi t)\) with respect to \(t\) involves using the chain rule. The chain rule states that if \(u(t)\) is a differentiable function, then the derivative of \(f(u(t))\) with respect to \(t\) is \(f'(u(t)) \cdot u'(t)\).
\[ a(t) = \pi \cos(\pi t) \]
Now that we have the expression for acceleration \(a(t)\), you can find the acceleration at \(t = 2\) by substituting \(t = 2\) into the expression:
\[ a(2) = \pi \cos(\pi \cdot 2) \]
\[ a(2) = \pi \cos(2\pi) \]
Since \(\cos(2\pi) = 1\), the acceleration at \(t = 2\) is:
\[ a(2) = \pi \cdot 1 = \pi \]
Therefore, the acceleration of the particle at \(t = 2\) is \(\pi\).
