Answer
a) 13 mi/h
b) 7 mi/h
Work Step by Step
a) $v_{avg} = \frac{s(3)-s(1)}{3-1} = \frac{(3*(3)^2+3)-(3*(1)^2+1)}{2} = \frac{30-4}{2} = 13 \text{ mi/h}$
b) The instantaneous velocity at $t=1$ is the derivative of $s$ at $t=1$:
$\frac{d}{dt} s = \frac{d}{dt} 3t^2+t = 6t+1 = 6(1) + 1 = 7 \text{ mi/h}$
