#### Answer

a)
$x_{0}=50 cm$
$v_{0} = 2 cm/s$
$a_{0}=-0.125cm/s^{2}$
b) $t=16s$
c) $t=32s$ after starting
d)
$t\approx 6.2s \Longrightarrow v(t\approx6.2s)\approx 1.22cm/s$
$t\approx 25.8s \Longrightarrow v(t\approx 25.8s)\approx -1.22cm/s$
$t\approx -4.4s \Longrightarrow v(t\approx -4.4s)\approx 2.55 cm/s$ (see the work step by step for more details)
$t\approx 36.4s \Longrightarrow v(t\approx 36.4s)\approx -2.55 cm/s$

#### Work Step by Step

\approxa) The equation for the turtle's position as a function of time is a quadratic equation. That means that the acceleration is constant. When the acceleration is constant, $x(t)$ is given by $x(t)=x_{0}+v_{0}t-\frac{a}{2}t^{2}$. We can make links between the given equation and this general form :
$x_{0}=50cm$
$v_{0}=2 cm/s$
$\frac{a}{2}=-0.0625 \Longleftrightarrow a=-0.125cm/s^{2}$
b) We find the time $t$ when $v(t)=0$. In such a motion, $v(t)=v_{0}+at$.
We can simply solve the equation :
$v(t)=2-0.125t=0$
$t=16s$
c)We find the time $t$ when $x(t)=x_{0}=50cm$. We can simply solve the equation :
$x(t)=50+2t-0.0625t^{2}=50$
$-0.0625t^{2}+2t=0$
$t(-0.0625t+2)=0$
We have 2 possibilities : $t=0s$ or $-0.0625t+2=0$. The question asks when the turtle $returns$ to its starting point. So $t>0$ and we eliminate the first option. The second option gives : $t=32s$.
d) The turtle is a distance of 10.0 cm from its starting point when $x=40cm$ and when $x=60cm$.
Let's first determine at which times $x=40cm$ :
$-0.0625t^{2}+2t+50=40$
$-0.0625t^{2}+2t+10=0$
$\Delta=6.5$
$t_{1}\approx-4.4s$ ; $t_{2}\approx 36.4s$
IF the given equation is defined for $t<0$, we can accept $t_{1}$ as an answer. The domain of the equation is not specified, but a negative time doesn't really makes sense in this case...
Let's now determine when $x=60cm$ :
$-0.0625t^{2}+2t+50=60$
$-0.0625t^{2}+2t-10=0$
$\Delta=1.5$
$t_{3}\approx6.2s$ ; $t_{4}\approx 25.8s$
We still have to calculate the velocities at each of those times...
$v(t)=v_{0}+at=2-0.125t$
$v(t_{1})\approx2.55cm/s$
$v(t_{2})\approx -2.55cm/s$
$v(t_{3})\approx 1.22cm/s$
$v(t_{4})\approx -1.22cm/s$