Find the velocity of a particle at \( t=4 \) given its position function. (position is in mi, time is in hr) \( s(t)=2 t^{3}+16 \sqrt{\mathrm{t}} \) \( v(4)=[?] \mathrm{mi} / \mathrm{hr} \)

To find the velocity of the particle at \( t=4 \), we first need to take the derivative of the position function \( s(t) \). The derivative \( v(t) \) is given by: \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(2t^3 + 16\sqrt{t}) \] Using the power rule and the chain rule, we differentiate: 1. \( \frac{d}{dt}(2t^3) = 6t^2 \) 2. For \( \frac{d}{dt}(16\sqrt{t}) \), we rewrite it as \( 16t^{1/2} \) and find \( \frac{d}{dt}(16t^{1/2}) = 8t^{-1/2} \). Thus, combining these results, we have: \[ v(t) = 6t^2 + \frac{8}{\sqrt{t}} \] Now we substitute \( t = 4 \): \[ v(4) = 6(4^2) + \frac{8}{\sqrt{4}} = 6(16) + \frac{8}{2} = 96 + 4 = 100 \] Therefore, the velocity of the particle at \( t = 4 \) is: \[ v(4) = 100 \, \mathrm{mi/hr} \]