The question is: A body moves along a straight line with acceleration $a$ in meters per second squared given by $a = \frac{7}{36}t$ where $t$ is the time in seconds. Initially the body is at rest at an origin $O$. The acceleration continues until $t = 6$, whereupon it ceases and the body is retarded to rest. During this retardation the acceleration is given by $a = -1/4t$. Find the value of $t$ when the body comes to rest and the displacement of the body from $O$ at that time.
I understand how to use integration to find velocity and displacement, I have written out the equation for velocity using integration but I am not sure what to do after this.
start from $a = \frac{7}{36}t \ $ like below $$a=\frac {dv(t)}{dt}= \frac{7}{36}t\\ \to dv(t)= \frac{7}{36}tdt\\\int_0^6dv(s)ds=\int_0^6\frac{7}{36}sds\\v(s)|_0^6=\frac{7}{36}\frac {s^2}{2}|_0^6\\v(6)-v(0)=3.5\\W.R.T. \ v(0)=0 \to v(6)=3.5 $$ for the second part $v(6)=3.5 \frac {m}{s}$ and now $$a=\frac {dv(t)}{dt}=\frac{-1}{4}t \ \frac ms\\ dv(t)=\frac{-1}{4}tdt\\ \int_6^{t}dv(s)=\int_6^{t}\frac{-1}{4}sds\\v({t})-v(6)=\frac {-1}{8}(s^2)|_6^t\\v(t)-3.5=\frac {-1}{8}(t^2-36)$$ if you want to find when $v(t)=0$ solve the equation $$v(t)=3.5+\frac {-1}{8}(t^2-36)=0 \to t=8 \ sec$$ at the end , a figure can achieve them