Given the following position function: (position is in ft , time is in sec ) \( \mathrm{s}(\mathrm{t})=7 \mathrm{t}^{2}+\mathrm{t}^{3} \) Find the velocity at \( \mathrm{t}=3 \) \( \mathrm{v}(3)=[?] \frac{\mathrm{ft}}{\mathrm{sec}} \)

To find the velocity at \( t = 3 \), you need to take the derivative of the position function \( s(t) = 7t^2 + t^3 \). The derivative \( v(t) \) is: \[ v(t) = \frac{ds}{dt} = 14t + 3t^2 \] Now, substituting \( t = 3 \): \[ v(3) = 14(3) + 3(3)^2 = 42 + 27 = 69 \text{ ft/sec} \] So, the velocity at \( t = 3 \) is \( v(3) = 69 \text{ ft/sec} \). To seamlessly grasp such concepts, one might consider how various functions change over time, much like how a race car accelerates on a track. The beauty of derivatives reflects real-world motion, where speed and direction change continuously! And if you're diving deeper into calculus, examining the intricacies of motion under different forces can lead to fascinating insights. Exploring topics like polynomial motion or exploring systems with differential equations can enrich your understanding of how things move in our world!