a) In order to graph acceleration vs time, it is necessary to calculate acceleration for four different time intervals. Use the following formula for the acceleration:
[tex]a=\frac{v-v_o}{t}[/tex]
where v is the final speed and vo is the initial speed.
Between t=0.0 and t=0.5 you have for a:
[tex]a=\frac{4-2}{0.5}\frac{m}{s^2}=4\frac{m}{s^2}[/tex]
Between t=0.5 and t=1.0:
[tex]a=\frac{4-4}{0.5}\frac{m}{s^2}=0\frac{m}{s^2}[/tex]
Between t=1.0 and t=2.0:
[tex]a=\frac{-2-4}{1.0}\frac{m}{s^2}=-6\frac{m}{s^2}[/tex]
Finally, between t=2.0 and t=2.5:
[tex]a=\frac{0-(-2)}{0.5}\frac{m}{s^2}=4\frac{m}{s^2}[/tex]
Then, with the previous values of the acceleration you have the following
a vs t graph:
b) In order to graph position vs time for the first second of motion, use the following formula for position x:
[tex]x=v_ot+\frac{1}{2}at^2[/tex]
Consider that between t=0 and t=0.5 the acceleration is 4m/s^2 and the initial speed is 2m/s, then, you have for this time interval:
[tex]\begin{gathered} x=(2\frac{m}{s})(0.5s)+\frac{1}{2}(4\frac{m}{s^2})(0.5s)^2 \\ x=1.5m \end{gathered}[/tex]
Between t=0.5 and t=1.0 the acceleration is zero, then, the speed of squirrel is constant (4m/s). The value of the final position is then:
[tex]x=vt=(4\frac{m}{s})(0.5s)=2m[/tex]
In order to graph position vs time, consider that in the first time interval you have a parabolla (there is an acceration) and in the second interval you have a line (constant speed):
