Question

The position of a passenger train that is traveling at an initial speed of 14 feet per second and continues to accelerate can be modeled by the function y = 14t2. A second train that is 1,200 feet ahead of the first train is traveling at a constant speed of 130 feet per second and can be modeled by the function y = 130t + 1200. Solve the system of equations. Which solution represents a viable time that the trains are side by side?

14 sec
15 sec
16 sec
17 sec

Answer 1

Answer: 15 seconds

Step-by-step explanation: Set both equations equal to each other to solve for the time when both trains would be side-by-side.

14t²=130t+1200

14t²-130t-1200=0

x=
`\frac{-b+\sqrt{b^(2)-4ac } }{2a} \\`and/or
`\frac{-b-\sqrt{b^(2)-4ac } }{2a} \\`

a=14, b=-130, c=-1200

x=
`\frac{-(-130)+\sqrt{(-130)^(2)-4(14)(-1200) } }{2(14)} \\`and/or
`\frac{-(-130)-\sqrt{(-130)^(2)-4(14)(-1200) } }{2(14)} \\`

x=
`\frac{130+\sqrt{290^(2) } }{28} \\` and/or
`\frac{130-\sqrt{290^(2) } }{28} \\`

`x_(1\\)`=
`(130+290)/(28)` and
`x_(2)`=
`-(40)/(7)`

`x_(1\\)`=15 seconds and
`x_(2)`=
`-(40)/(7)`

The value
`x_(2)` can be discarded because it is negative. Both trains are parallel at 15 seconds.