Answer
The speed at the end of the $~~1.6~s~~$ is $~~8.8~m/s$
Work Step by Step
We can find the acceleration:
$x = \frac{1}{2}at^2$
$a = \frac{2x}{t^2}$
$a = \frac{(2)(7.0~m)}{(1.6~s)^2}$
$a = 5.47~m/s^2$
We can find the speed at the end of the $1.6~s$:
$v_f = v_0+at$
$v_f = 0+at$
$v_f = (5.47~m/s^2)(1.6~s)$
$v_f = 8.8~m/s$
The speed at the end of the $~~1.6~s~~$ is $~~8.8~m/s$
