Train/Car acceleration/speed problem

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A train moves at a constant speed of 24 m/s while a car, initially 49 m behind, travels at 37 m/s and accelerates at 4 m/s². The car's final velocity when it passes the train is calculated using the equation 2aΔx = v² - Vo², resulting in approximately 41.96 m/s. The challenge lies in understanding how the train's speed influences the car's speed as it overtakes the train. The car must exceed the train's speed to pass, necessitating a clear approach to incorporate the train's constant speed into the calculations.
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Homework Statement


A train is moving parallel and adjacent to a highway with a constant speed of 24 m/s. Initially a car is 49 m behind the train, taveling in the same direction as the train at 37 m/s, and accelerating at 4 m/s/s.

What is the spped of the car just as it passes the train? Answer in units of m/s.



Homework Equations



I used 2aΔx=v^2-Vo^2


The Attempt at a Solution



I was solving for the car to get its final velocity (v^2) when it passes the train:
2aΔx=v^2-Vo^2
2(4m/s/s)(49m)=V^2-(37m/s)^2
392 m^s/s^2=V^2-1369 m^2/s^2
1761 m^2/s^2=V^2
41.96427 m/s=V (for the car)

my problem is that I don't know how to accommodate for the trains speed and how it effects the speed of the car as it passes. i know that the car will have to go faster but I just don't know how to approach the problem
 
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