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