Physics acceleration and velocity homework question?

We need to write two kinematic equations one for brown road runner and another one for black road runner.

Brown road runner.

We note that brown road runner is standing still. It accelerate at a rate #=2.75 ms^-2# and attempts to catch up with the black road runner.
Let us assume that it is able to catch up the other road runner after time #=t#
Distance traveled by brown road runner can be found by the kinematic expression
#s=ut+1/2at^2#
Inserting given numbers we get
#s=0xxt+1/2xx2.75t^2#
#=>s=1.375t^2# ......(2)

Black road runner.

We see that this road runner travels at a constant speed #=9.25ms^-1#
This road runner also travels distance #s# in time #t# when brown catches up to black road runner.
Kinematic expression for this road runner is
#"Distance traveled"="Speed"xx"Time"#
#s=9.25t# .....(2)
To find distance #s# we eliminate #t# from equation (1) with the help of (2). From (2)
#t=s/9.25#
Inserting this value in (1) we get
#s=1.375(s/9.25)^2#
#=>1.375/(9.25)^2s^2-s=0#
#=>s(1.375/(9.25)^2s-1)=0#
Two roots of this quadratic are
#s=0#
#1.375/(9.25)^2s-1=0#

Ignoring first root as it represents when black road runner first time overtook brown road runner. We get
#s=1xx(9.25)^2/1.375#
#s=62.23m#, rounded to two decimal places.