At the instant a motorbike starts from rest in a given direction, a car overtakes the bike, both moving in the same direction as shown in figure. The speed-time graphs for motor bike and car are represented by OAB and CD respectively.

Then,

Distance travelled by motor bike at t = 18 s
$s_1=\frac{1}{2}\times 18\times 60=540\mathrm{m}$
Distance travelled by the car at t = 18 s
S₂ = 18 x 40 =720m
The separation between motor bike and car at t = 18 s is 720 - 540 = 180 m
Let the separation between them decreases to zero at time t beyond 18 s. Hence
${\mathrm{s}}_{\mathrm{bike} }=540+60\mathrm{t} \mathrm{and} {\mathrm{s}}_{\mathrm{car} }=720+40\mathrm{t}$
$\therefore {\mathrm{s}}_{\mathrm{car} }-{\mathrm{s}}_{\mathrm{bike}}=0$
Therefore,720 + 40 t = 540 + 60 t
Solving we get, t = 9 s beyond 18 s
total time = 18 + 9 = 27 s from the start.
Distance travelled by both s_bike + s_car = 1080 m