Question The velocity of the center of mass of a system of particles changes as \( v=4.5 t+2.4 t^{2}+1.1 t^{3} \), where \( v \) is in meters per second. If the system starts from rest at \( t=0 \), what is its acceleration at \( t=3.0 \mathrm{~s} \) ? \[ \begin{array}{l} 7.1 \mathrm{~m} / \mathrm{s}^{2} \\ 14 \mathrm{~m} / \mathrm{s}^{2} \\ 20 \mathrm{~m} / The velocity of the center of mass of a system of particles changes as \( v=4.5 t+2.4 t^{2}+1.1 t^{3} \), where \( v \) is in meters per second. If the system starts from rest at \( t=0 \), what is its acceleration at \( t=3.0 \mathrm{~s} \) ? \[ \begin{array}{l} 7.1 \mathrm{~m} / \mathrm{s}^{2} \\ 14 \mathrm{~m} / \mathrm{s}^{2} \\ 20 \mathrm{~m} / \mathrm{s}^{2} \\ 49 \mathrm{~m} / \mathrm{s}^{2} \\ 65 \mathrm{~m} / \mathrm{s}^{2} \end{array} \]

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Transcribed Image Text: The velocity of the center of mass of a system of particles changes as \( v=4.5 t+2.4 t^{2}+1.1 t^{3} \), where \( v \) is in meters per second. If the system starts from rest at \( t=0 \), what is its acceleration at \( t=3.0 \mathrm{~s} \) ? \[ \begin{array}{l} 7.1 \mathrm{~m} / \mathrm{s}^{2} \\ 14 \mathrm{~m} / \mathrm{s}^{2} \\ 20 \mathrm{~m} / \mathrm{s}^{2} \\ 49 \mathrm{~m} / \mathrm{s}^{2} \\ 65 \mathrm{~m} / \mathrm{s}^{2} \end{array} \]
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Transcribed Image Text: The velocity of the center of mass of a system of particles changes as \( v=4.5 t+2.4 t^{2}+1.1 t^{3} \), where \( v \) is in meters per second. If the system starts from rest at \( t=0 \), what is its acceleration at \( t=3.0 \mathrm{~s} \) ? \[ \begin{array}{l} 7.1 \mathrm{~m} / \mathrm{s}^{2} \\ 14 \mathrm{~m} / \mathrm{s}^{2} \\ 20 \mathrm{~m} / \mathrm{s}^{2} \\ 49 \mathrm{~m} / \mathrm{s}^{2} \\ 65 \mathrm{~m} / \mathrm{s}^{2} \end{array} \]
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OAOPP7

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/3Given \( \begin{align*} \mathrm{{v}} &= \mathrm{{4.5}{t}+{2.4}{t}^{{{2}}}+{1.1}{t}^{{{3}}}} \end{align*} \)we have to calculate acceleration at t = 3.0 s Explanation:here of use differentiation to find acceleration at 3 s .Step2/3since acceleration: change in velocity with respect to time i.e a = dv/dt\( \begin{align*} \mathrm{{a}} &= \mathrm{\frac{{{d}{\left({4.5}{t}+{2.4}{t}^{{{2}}}+{1.1}{t}^{{{3}}}\right)}}}{{{\left.{d}{t}\right.}}}} \\[3pt]\mathrm{{a}} &= \ ... See the full answer