At the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed time graphs for motor bike and car are represented by $OAB \& CD$ respectively. Then

Distance travelled by motor bike at $t= 18 s$

$S_{bike}=S_1=\frac{1}{2}\left(18\right)\left(60\right)=540m$

Distance travelled by car at $t=18 s$

$S_{car}=S_2=\left(18\right)\left(40\right)=720 m$

Therefore, seperation between them at $t=18s$ is $180m$. Let seperation between them decreases to zero at time $t$ beyond $18s$.

Hence $S_{bike}=540+60t \& S_{car}=720+40t$

$S_{car}-S_{bike}=0\Rightarrow 720+40t=540+60t$

$t=\left(18+9\right)s=27s$ from start and distance travelled by both is $S_{bike}=S_{car}=1080 m$