Answer
(a) The velocity is zero again at $t = 20~s$
(b) $x = 667~m$
Work Step by Step
(a) $v(t) = v_0+\int_{0}^{t}~a(t)~dt$
$v(t) = 0+\int_{0}^{t}~(10-t)~dt$
$v(t) = 10t-\frac{1}{2}t^2$
We can then find $t$ when $v(t) = 0$;
$10t-\frac{1}{2}t^2 = 0$
$(10-\frac{1}{2}t)~t = 0$
$t = 0$ or $t = 20$
Therefore, velocity is zero again at $t = 20~s$.
(b) $x(t) = x_0+\int_{0}^{t}~v(t)~dt$
$x(t) = 0+\int_{0}^{t}~10t-\frac{1}{2}t^2~dt$
$x(t) = 5t^2-\frac{1}{6}t^3$
We can then find $x(t)$ at $t = 20 ~s$;
$x = 5t^2-\frac{1}{6}t^3$
$x = (5)(20)^2-\frac{1}{6}(20)^3$
$x = 667~m$
