Car a, traveling at a constant 10 m/s, crosses the start line 3 seconds before car b, who is traveling at a constant 15 m/s. at what time and distance, respectively, will the two cars be at the same distance from the start line?

The distance between Car A and car B,When Car A crosses the start line is:
distance =speed car B* time
distance=(15 m/s)(3 s)=45 m
 
Distance traveled by car A =x,  (when the car B is at the same distance from the start line)
time of car A=t

x=10 m/st  ⇒    x=10t    (1)

Distance traveled by car B=x
time of car B=t-3

x=15(t-3)  (2)

With the equations (1) and (2) we make a system of equations:

x=10t
x=15(t-3)

We solve this system of equations:

10t=15(t-3)
10t=15t-45
-5t=-45
t=-45 / -5
t=9

t-3=9-3=6

x=10 t=10 (9)=90

The time would be 9 seconds for Car A and 6 seconds for car B and the distance would be 90 meters.