If the velocity of an object is given by \( v(t) = 3t^2 + 2t \), what is its acceleration at \( t = 4 \) seconds?

["### Understanding Acceleration: Finding It from Velocity at ( t = 4 )", "When analyzing motion, one of the fundamental concepts in physics is acceleration—the rate at which an object’s velocity changes over time. If you’re given the velocity function ( v(t) = 3t^2 + 2t ), determining the acceleration at a specific time, such as ( t = 4 ) seconds, becomes a straightforward calculus problem.", "#### What Is Acceleration?", "Acceleration is the derivative of velocity with respect to time. Mathematically, if ( v(t) ) represents velocity, then acceleration ( a(t) ) is given by:", "[\na(t) = \frac{dv}{dt}\n]", "This derivative tells us how quickly the velocity is changing at any instant.", "#### Step-by-Step: Calculating Acceleration", "Given the velocity function:", "[\nv(t) = 3t^2 + 2t\n]", "To find acceleration, take the derivative of ( v(t) ) with respect to ( t ):", "1. Differentiate ( 3t^2 ):\n [\n \frac{d}{dt}(3t^2) = 6t\n ]", "2. Differentiate ( 2t ):\n [\n \frac{d}{dt}(2t) = 2\n ]", "Adding these together, the acceleration function is:", "[\na(t) = 6t + 2\n]", "#### Evaluating Acceleration at ( t = 4 )", "Now, substitute ( t = 4 ) seconds into the acceleration expression:", "[\na(4) = 6(4) + 2 = 24 + 2 = 26 , \ ext{m/s}^2\n]", "### Conclusion: The Acceleration at ( t = 4 ) seconds", "At ( t = 4 ) seconds, the acceleration of the object is:", "[\n\boxed{26 , \ ext{m/s}^2}\n]", "This positive acceleration indicates the object is speeding up in the positive direction of motion at that moment. Whether you’re studying for physics class, planning engineering applications, or curious about motion, understanding how to derive acceleration from velocity is essential—and now you’ve learned how to compute it precisely!"]