Answer
(a) $t = 2$ or $t=5$
(b) At t = 2 s:
$a = -3~m/s^2$
At t = 5 s:
$a = 3~m/s^2$
Work Step by Step
(a) $v(t) = t^2-7t+10$
We can find the turning points when $v(t) = 0$;
$t^2-7t+10 = 0$
$(t-5)(t-2) = 0$
$t = 2$ or $t=5$
(b) $a(t) = \frac{dv}{dt} = 2t-7$
At t = 2 s:
$a = (2)(2)-7$
$a = -3~m/s^2$
At t = 5 s:
$a = (2)(5)-7$
$a = 3~m/s^2$
