Answer
(a) $ 45 \frac{2}{3} km/h$
(b) $5.2s$
Work Step by Step
(a) $v_{ave} = \frac{1}{12} \int^{12}_0 v(t) dt = \frac{1}{12} I$ Use the Midpoint Rule with $n = 3$ and $\Delta t = 4$ to estimate I.
$I \approx M_3 = 4[v(2) +v(6) +v(10)] = 4[21+50+66] = 548$. Thus, $v_{ave} \approx \frac{1}{12} \times 548 = 45 \frac{2}{3} km/h$
(b) Estimating from the graph, $v(t) = 45 \frac{2}{3}$ when $t \approx 5.2s$