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

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