Answer
(a) After t = 0, the particle first reaches a turning point at t = 1 s.
(b) $a = -6.28~m/s^2$
Work Step by Step
(a) $v(t) = (2.0~m/s)~sin(\pi~t)$
We can find turning points when $v(t) = 0$. Note that $v(t) = 0$ when $t = 0, 1, 2, 3, 4,...$
After t = 0, the particle first reaches a turning point at t = 1 s.
(b) $a(t) = \frac{dv}{dt}$
$a(t) = (2.0~m/s)(\pi)~cos(\pi~t)$
We can then find the acceleration at t = 1 s;
$a = (2.0~m/s)(\pi)~cos(\pi~t)$
$a = (2.0~m/s)(\pi)~cos(\pi)$
$a = (2.0~m/s)(\pi)(-1)$
$a = -6.28~m/s^2$
