Answer
$320$ m
$-32$ m/s
Work Step by Step
Given the acceleration of $-1.6 m/s^2$ on the moon, the acceleration function must be:
$a(t) = -1.6$
To find the velocity function, integrate the acceleration function.
$\int a(t) = v(t) = -1.6t + v_0$
Since the rock is dropped, the initial velocity, $v_0$, is zero.
$v(t) = -1.6t$
To find the distance that the rock fell, integrate the velocity function from $t = 0$ to $t = 20$.
$\int_0^{20}v(t) = (-0.8t^2)|_0^{20}$
$(-0.8t^2)|_0^{20} = (-0.8(20)^2) - (-0.8(0)^2) = -320$
So, the rock fell $320$ m.
To find the velocity when the rock hits the ground, find $v(20)$.
$v(20) = -1.6(20)$
$v(20) = -32$
So, the rock was travelling at $-32$ m/s when it hit the ground.
