Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 682: 57

$\sqrt{293}$

Recall: The speed of the particle is formulated by $v(t)=\sqrt{(dx/dt)^2+(dy/dt)^2}$. Find $dx/dt$ and $dy/dt$: $dx/dt=2$ and $dy/dt=4t-3$ Evaluate $v(t)$ at $t=5$: $v(t)=\sqrt{2^2+(4t-3)^2}$ $v(5)=\sqrt{2^2+(4\cdot 5-3)^2}$ $v(5)=\sqrt{4+17^2}$ $v(5)=\sqrt{293}$ Thus, the speed of the particle at time $t=5$ is $\sqrt{293}$.