Answer :
Given that
[tex]v = (t^{2} - 4t +3)[/tex]
the time interval is
[tex]0\leq t\leq 4s[/tex]
We calculate the time
[tex]v =t^{2} -4t +3[/tex]
[tex]a = \dfrac{dv}{dt} = \dfrac{d}{dt}(t^{2} -4t +3)[/tex]
[tex]a = 2t - 4[/tex]
[tex]t = 2s[/tex]
Now, the graph of v-t for time interval [tex]0\leq t\leq 4s[/tex]
[tex]v =t^{2} -4t +3[/tex]
at [tex]t = 0\ s , v = 0 - 0 + 3 = 3 m/s[/tex]
at [tex]t = 2\ s , v = 4 - 8 +3 = -1 m/s[/tex]
at [tex]t = 4\ s , v = 16 - 16 +3 = 3 m/s[/tex]
It is show in figure 1
Now, the graph of a-t for time interval
[tex]v =t^{2} -4t +3[/tex][tex]a = \dfrac{dv}{dt} = \dfrac{d}{dt}(t^{2} -4t +3)[/tex]
[tex]a = 2t - 4[/tex]
at [tex]t = 0\ s , a = - 4 m/s^{2}[/tex]
at [tex]t = 2\ s , a = 0 m/s^{2}[/tex]
at [tex]t = 4\ s , a = 4 m/s^{2}[/tex]
It is show in figure 2
Hence, this is the required solution
