A mass hanging from a spring is set in motion and its ensuing velocity is given by \( v(t)=-3 \pi \sin \pi t \) for \( t \geq 0 \). Assume that the positive direction is upward and \( s(0)=3 \). a. Determine the position function for \( t \geq 0 \). b. Graph the position function on the interval \( [0,3] \). c. At what times does the mass reach its lowest point the first three times? d. At what times does the mass reach its highest point the first three times?
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To find the position function, we first need to integrate the velocity function. The velocity function is given by \( v(t) = -3\pi \sin(\pi t) \). ### a. Determine the position function for \( t \geq 0 \). The position function \( s(t) \) can be found by integrating the velocity function: \[ s(t) = \int v(t) \, dt = \int -3\pi \sin(\pi t) \, dt \] Using the integral of \(\sin\), we have: \[ s(t) = 3\cos(\pi t) + C \] To find the constant \( C \), we use the initial condition \( s(0) = 3 \): \[ s(0) = 3\cos(0) + C = 3 \implies C = 0 \] Thus, the position function is: \[ s(t) = 3\cos(\pi t) \] ### b. Graph the position function on the interval \( [0,3] \). The graph of \( s(t) = 3\cos(\pi t) \) will oscillate between its maximum value, 3 (when \( t = 0 \), \( 2 \), ...), and its minimum value, -3 (when \( t = 1 \), \( 3 \), ...). This function has a period of 2.5 seconds, meaning it will reach its highest and lowest points twice in the interval [0, 3]. In the first cycle from \( t = 0 \) to \( t = 2 \), the graph will start at \( s(0) = 3 \), decrease to the lowest point \( s(1) = -3 \), and return to the maximum at \( s(2) = 3 \). In this interval, you will see a complete oscillation, starting from the maximum, reaching the minimum, and coming back up. ### c. At what times does the mass reach its lowest point the first three times? To find the times \( t \) when the mass reaches its lowest point, we set \( s(t) = -3 \): \[ 3\cos(\pi t) = -3 \implies \cos(\pi t) = -1 \] This occurs when \( \pi t = (2n + 1)\pi \) for integers \( n \), hence: \[ t = 2n + 1 \quad \text{where } n = 0, 1, 2, \ldots \] The lowest point occurs at: - For \( n = 0 \): \( t = 1 \) - For \( n = 1 \): \( t = 3 \) - For \( n = 2 \): \( t = 5 \) So the first three times are \( t = 1, 3, 5 \). ### d. At what times does the mass reach its highest point the first three times? To find the times \( t \) when the mass reaches its highest point, we set \( s(t) = 3 \): \[ 3\cos(\pi t) = 3 \implies \cos(\pi t) = 1 \] This occurs when \( \pi t = 2n\pi \), leading to: \[ t = 2n \quad \text{where } n = 0, 1, 2, \ldots \] The highest point occurs at: - For \( n = 0 \): \( t = 0 \) - For \( n = 1 \): \( t = 2 \) - For \( n = 2 \): \( t = 4 \) So the first three times are \( t = 0, 2, 4 \).